3.3 Trigonometric Substitution

1. Trigonometric Substitution

Title

Calculus II · Section 3.3

Turning a square root into a perfect square

2. What this lesson gives you

Objectives

Roots of a sum or difference of two squares resist every technique so far. One idea handles all of them: replace x by a trigonometric function chosen so that a Pythagorean identity turns the radicand into a single square.

• Match a radical to one of the three standard forms and choose the substitution
• Explain why the substitution clears the root, from a Pythagorean identity and a range for the angle
• Draw the reference triangle and convert an answer back to x
• Complete the square to bring a general quadratic into a standard form
• Handle definite integrals by converting the limits, and recognise when a plain substitution is quicker

OpenStax Calculus Volume 2, §3.3 Trigonometric Substitution §3.3, pp. 285-297 — learning objective 3.3.1

So far you have two big tools for integrals: ordinary substitution, which undoes the chain rule, and integration by parts, which undoes the product rule. Section 3.2 added a third, the trigonometric integrals, where identities trade one power of sine or secant for another. This lesson uses that third tool to crack a family of integrals that none of the others can touch: anything containing the square root of a sum or a difference of two squares.

The idea is a single move. Instead of fighting the root, you change variables so that the expression under it becomes one perfect square, and then the root simply disappears. The right change of variable is always a trigonometric function, because the Pythagorean identities are exactly the statements that turn two squares into one.

By the end you should be able to look at a root, name its form, write down the substitution without hesitation, finish the integral, and convert back, either with a triangle or by converting the limits.

3. Before anything new: the identities you will lean on

Warm-up

Write your answers before revealing them. If any of the three identities takes you more than a few seconds, rebuild it: divide sine squared plus cosine squared equals one by cosine squared and you get the tangent and secant version in one line.

Look at the shape of each identity rather than its letters. Each one says that a sum or a difference of two squares is a single square. One minus a square of sine is a square of cosine. One plus a square of tangent is a square of secant. A square of secant minus one is a square of tangent. Those three shapes will match the three kinds of root in this lesson, one to one.

The antiderivatives matter too. Almost every example today ends in cosine squared, secant, or secant cubed, so having those results at your fingertips means the new part of each problem is only the substitution itself.

4. The sine substitution

Section

Part 1

5. A root no earlier technique opens

Concept

\[ \int \sqrt{9 - x^2}\,dx \]

Try u equal to 9 minus x squared: its derivative needs a factor of x that is not there. Integration by parts has nothing useful to differentiate. And yet the graph is a semicircle, so every definite integral of it is an area of part of a disc.

OpenStax Calculus Volume 2, §3.3 Trigonometric Substitution §3.3, p. 285 — introduction

Try the obvious things on this integral and watch them fail. Ordinary substitution with u equal to 9 minus x squared needs a factor of x in the integrand to pair with du, and there is none. Integration by parts has no product worth splitting. The integrand is not a derivative you recognise.

Now look at the picture. The graph of the root of 9 minus x squared is the upper half of a circle of radius 3, so the shaded region is a slice of a disc. Its area is certainly a definite, finite number, which tells you an antiderivative exists; the techniques so far just cannot find it.

When the geometry is circular, the natural language is angles. That is the hint the next slide follows up.

6. The circle suggests the substitution

Concept

\[ x = 3\sin\theta \quad\Longrightarrow\quad \sqrt{9 - x^2} = 3\cos\theta \]

Describe the point by its angle instead of by x, and the awkward root becomes a plain cosine. The new variable is an angle, and the integral becomes a trigonometric integral of the kind Section 3.2 handles.

Pick a point on the circle and describe it by the angle theta its radius makes with the vertical axis. Right-triangle trigonometry gives its horizontal coordinate as 3 times the sine of theta and its height as 3 times the cosine of theta.

But the height of the point on this circle is exactly the root of 9 minus x squared. So if you write x as 3 sine theta, the root becomes 3 cosine theta, with no square root left anywhere. That is the whole trick, seen geometrically.

Notice the angle is measured from the vertical, which is why x comes out as a sine rather than a cosine. Either choice would work, but sine is the convention the book uses, and it makes the range of theta the familiar one for the inverse sine.

7. Why x = a sin theta clears the root

Concept

Replace x by a sine of theta, with theta between minus and plus a right angle, and simplify one move at a time.

\[ \sqrt{a^2 - x^2} = \sqrt{a^2 - a^2\sin^2\theta} \]

\[ = \sqrt{a^2\left(1 - \sin^2\theta\right)} \]

\[ = \sqrt{a^2\cos^2\theta} = |a\cos\theta| \]

\[ = a\cos\theta \qquad \left(a > 0,\; \cos\theta \ge 0 \text{ for } -\tfrac{\pi}{2} \le \theta \le \tfrac{\pi}{2}\right) \]

The identity turns two squares into one, and the square root of a square is an absolute value. The range for theta is what lets you drop the bars.

OpenStax Calculus Volume 2, §3.3 Trigonometric Substitution §3.3, p. 285 — the derivation before Figure 3.4

This is the algebra behind the picture, and it is worth being able to reproduce. Substitute, factor the a squared out of both terms, and the Pythagorean identity turns one minus sine squared into cosine squared. The root of a square is not the thing itself but its absolute value, and that is the step people skip.

The absolute value disappears only because theta is restricted to the interval from minus a right angle to plus a right angle, where cosine is never negative, and because a itself is positive. Without that restriction the last line would be false for some angles, as a later slide makes you prove.

The same three moves, substitute, factor, apply the identity, will clear the other two kinds of root. Only the identity changes.

8. Reading the sine substitution

Notation

OpenStax Calculus Volume 2, §3.3 Trigonometric Substitution §3.3, p. 286 — Problem-Solving Strategy, step 2

Step through the notes one at a time. The first line is a choice you make; the other two are consequences you must carry out every time. In particular, the differential is not optional: an integral in theta needs d theta, and dx is not d theta.

The range note explains why the substitution is legitimate. The root only exists for x between minus a and a, and a times sine of theta covers exactly that interval once as theta runs over its range. So every x corresponds to exactly one theta, and you can always get back.

The last note is how you get back when theta survives to the final answer on its own: theta is the inverse sine of x over a. Everything else, sine, cosine, tangent of theta, comes from the reference triangle two slides from now.

9. Why theta stays between minus and plus a right angle

Picture it

On the highlighted strip, sine covers every value between minus 1 and 1 exactly once, so every x between minus a and a has exactly one theta. On the same strip cosine is never negative, so the root equals a cos theta rather than its absolute value.

The highlighted strip is the interval for theta. Follow the sine curve across it: it climbs from minus 1 to 1 without ever turning back, so each x between minus a and a is reached exactly once. That is what lets the substitution be undone with an inverse sine.

Now follow the cosine curve over the same strip. It rises from zero to one and back to zero, and it is never negative there. Outside the strip, in the regions labelled in red, cosine dips below zero. If theta were allowed out there, the root, which is never negative, could not equal a times cosine of theta.

So the restriction is not fussiness. It is the reason the substitution is one-to-one and the reason the absolute value can be dropped.

10. The reference triangle: from theta back to x

Concept

\[ \sin\theta = \frac{x}{a}, \quad \cos\theta = \frac{\sqrt{a^2 - x^2}}{a}, \quad \tan\theta = \frac{x}{\sqrt{a^2 - x^2}} \]

\[ \theta = \sin^{-1}\left(\frac{x}{a}\right) \]

The answer comes out in theta; the question was asked in x. Read every other trigonometric function off the triangle as a ratio of two sides.

OpenStax Calculus Volume 2, §3.3 Trigonometric Substitution §3.3, p. 286 — Figure 3.4

After you integrate, the answer is written in theta, but the question was in x. The reference triangle is how you translate. The substitution says sine theta is x over a, so draw a right triangle with x opposite the angle and a on the hypotenuse.

Pythagoras gives the third side, the adjacent one, as the root of a squared minus x squared. Now any trigonometric function of theta is a ratio of two sides you can read off: cosine is adjacent over hypotenuse, tangent is opposite over adjacent, and so on.

The triangle is drawn for positive x, but the book notes that its ratios give the correct values for every theta in the range, negative ones included. And if theta itself appears in the answer, use the inverse sine.

11. Example 3.21: the root of nine minus x squared

Worked example

\[ \int \sqrt{9 - x^2}\,dx \]

\[ x = 3\sin\theta, \quad dx = 3\cos\theta\,d\theta \]

\[ \sqrt{9 - 9\sin^2\theta} = \sqrt{9\cos^2\theta} = 3\cos\theta \]

\[ \int 3\cos\theta \cdot 3\cos\theta\,d\theta = \int 9\cos^2\theta\,d\theta \]

\[ = \int 9\left(\tfrac12 + \tfrac12\cos 2\theta\right)d\theta = \tfrac92\theta + \tfrac94\sin 2\theta + C \]

\[ = \tfrac92\theta + \tfrac92\sin\theta\cos\theta + C \]

\[ \theta = \sin^{-1}\tfrac{x}{3}, \quad \sin\theta = \tfrac{x}{3}, \quad \cos\theta = \tfrac{\sqrt{9 - x^2}}{3} \]

\[ = \tfrac92\sin^{-1}\left(\tfrac{x}{3}\right) + \tfrac12 x\sqrt{9 - x^2} + C \]

\[ \tfrac{9}{2\sqrt{9 - x^2}} + \tfrac{\sqrt{9 - x^2}}{2} - \tfrac{x^2}{2\sqrt{9 - x^2}} = \tfrac{9 + (9 - x^2) - x^2}{2\sqrt{9 - x^2}} = \sqrt{9 - x^2} \]

OpenStax Calculus Volume 2, §3.3 Trigonometric Substitution §3.3, pp. 286-287 — Example 3.21

Here is the opening integral done completely. The substitution has three parts, x, dx and the root, and once they are in place the integrand is 9 cosine squared, which you integrate with the half-angle identity exactly as in Section 3.2.

The step people find awkward is getting back. The integral produced sine of 2 theta, and the triangle only tells you about theta. So use the double-angle formula first, sine of 2 theta equals 2 sine theta cosine theta, and only then read sine and cosine off the triangle. The theta on its own becomes an inverse sine of x over 3.

Then check. Differentiating the answer gives three terms; over the common denominator the numerator is 18 minus 2 x squared, which is twice 9 minus x squared, and that cancels down to the root you started with. When you differentiate an inverse sine, remember the chain rule factor of one third.

12. Slide a: the area under a semicircle

Tweak it

The curve is the antiderivative from Example 3.21, generalised to radius a and started at zero, so its height at any x is the area under the semicircle from 0 to x. Start with a equal to 3 and read the height at x equal to 3. You should see about 7.07, which is 9 pi over 4.

Now slide a. The height at x equal to a is always a quarter of the disc, pi a squared over 4, because the region from 0 to a is a quarter of the disc. The curve is steepest in the middle, where the semicircle is tallest, and flat at the ends, where the semicircle meets the axis.

Beyond x equal to a the root is not defined, so the curve is drawn flat. The picture is a reminder that the complicated formula with an inverse sine and a root is just the area of a slice of a disc.

13. Example 3.22: a root over x

Worked example

\[ \int \frac{\sqrt{4 - x^2}}{x}\,dx \]

\[ x = 2\sin\theta, \quad dx = 2\cos\theta\,d\theta, \quad \sqrt{4 - x^2} = 2\cos\theta \]

\[ \int \frac{2\cos\theta}{2\sin\theta}\,2\cos\theta\,d\theta = \int \frac{2\cos^2\theta}{\sin\theta}\,d\theta \]

\[ = \int \frac{2(1 - \sin^2\theta)}{\sin\theta}\,d\theta \]

\[ = \int \left(2\csc\theta - 2\sin\theta\right)d\theta \]

\[ = 2\ln|\csc\theta - \cot\theta| + 2\cos\theta + C \]

\[ \csc\theta = \tfrac{2}{x}, \quad \cot\theta = \tfrac{\sqrt{4 - x^2}}{x}, \quad \cos\theta = \tfrac{\sqrt{4 - x^2}}{2} \]

\[ F(x) = 2\ln\left|\frac{2 - \sqrt{4 - x^2}}{x}\right| + \sqrt{4 - x^2} + C \]

\[ F(1.5) - F(0.5) = -0.26786 + 2.19038 = 1.92253 \]

\[ \int_{0.5}^{1.5} \frac{\sqrt{4 - x^2}}{x}\,dx \approx 1.92253 \quad (\text{Simpson}, n = 2000) \]

OpenStax Calculus Volume 2, §3.3 Trigonometric Substitution §3.3, pp. 287-288 — Example 3.22

The substitution is the same kind, with a equal to 2, but the x in the denominator changes what happens after. Once everything is in theta you have cosine squared over sine, and the move is to trade the cosine squared for one minus sine squared so the fraction splits into cosecant minus sine.

The antiderivative of cosecant is the logarithm of cosecant minus cotangent. Then the triangle, with opposite x and hypotenuse 2, turns cosecant into 2 over x and cotangent into the root over x, and the two combine into one fraction inside the logarithm.

Differentiating this answer by hand is long, so the check here is numerical and just as honest: the change in the antiderivative between one half and three halves matches Simpson's rule applied to the original integrand to five decimal places. If a sign or a factor had slipped anywhere, those numbers would disagree.

14. Why the root keeps coming back

Intuition

\[ \text{hypotenuse } a, \;\text{ opposite } x \;\Longrightarrow\; \text{ adjacent } \sqrt{a^2 - x^2} \]

Both answers so far contain the very root the substitution removed. That is no accident: the root is a side of the reference triangle, so any trigonometric function that uses that side brings it back.

Use it as a sanity check. An answer to a root integral that has no root and no inverse trigonometric function in it deserves a second look.

Look back at both answers. Example 3.21 contains the root of 9 minus x squared and Example 3.22 contains the root of 4 minus x squared. You removed the root at the start, and it returned at the end.

The reason is the triangle. The substitution fixes two sides, and Pythagoras makes the third side equal to the root. So any trigonometric function that uses that side, cosine, tangent, cotangent, secant, brings the root back into the answer when you convert.

You can turn this into a quick plausibility check. If your final answer to an integral with a root in it contains neither that root nor an inverse trigonometric function, something has probably gone wrong in the conversion.

15. Trap: substituting for x but not for dx

Trap

16. Break the rule: an angle outside the range

Counterexample

\[ \sqrt{9 - x^2} = 3\cos\theta \quad \text{when } x = 3\sin\theta \]

Write your counterexample before revealing. Any angle where cosine is negative will do, and pi is the simplest.

At theta equal to pi, x is 3 sine pi, which is zero, and the root of 9 minus zero is 3. But 3 cosine pi is minus 3. The claimed equation fails, and it fails because a square root is never negative while cosine can be.

The restriction to the interval from minus a right angle to plus a right angle does two jobs at once: it makes sine one-to-one, so you can invert it, and it keeps cosine non-negative, so the absolute value from the square root can be dropped. Every trigonometric substitution in this section comes with such an interval, chosen for the same two reasons.

17. Example 3.23, method 1: an ordinary substitution

Worked example

\[ \int x^3\sqrt{1 - x^2}\,dx \]

\[ u = 1 - x^2, \quad du = -2x\,dx, \quad x^2 = 1 - u \]

\[ = -\tfrac12\int x^2\sqrt{1 - x^2}\,(-2x\,dx) \]

\[ = -\tfrac12\int (1 - u)\sqrt{u}\,du \]

\[ = -\tfrac12\int \left(u^{1/2} - u^{3/2}\right)du \]

\[ = -\tfrac12\left(\tfrac23u^{3/2} - \tfrac25u^{5/2}\right) + C \]

\[ = -\tfrac13\left(1 - x^2\right)^{3/2} + \tfrac15\left(1 - x^2\right)^{5/2} + C \]

\[ x(1 - x^2)^{1/2} - x(1 - x^2)^{3/2} = x\sqrt{1 - x^2}\left[1 - (1 - x^2)\right] = x^3\sqrt{1 - x^2} \]

OpenStax Calculus Volume 2, §3.3 Trigonometric Substitution §3.3, pp. 288-289 — Example 3.23, method 1

The first step of every strategy in this section is to check for something easier, and this integral shows why. There is a spare odd power of x outside the root, so u equal to 1 minus x squared works: one factor of x pairs with dx to make du, and the remaining x squared is 1 minus u.

After that it is only algebra: expand 1 minus u times the root of u into two powers, integrate each with the power rule, and put 1 minus x squared back in for u.

The check differentiates each term with the chain rule. Both terms share a factor of x times the root, and what is left in the bracket is 1 minus 1 minus x squared, which is x squared. So the derivative is x cubed times the root, exactly the integrand.

18. Example 3.23, method 2: the sine substitution

Worked example

\[ \int x^3\sqrt{1 - x^2}\,dx \]

\[ x = \sin\theta, \quad dx = \cos\theta\,d\theta, \quad \sqrt{1 - x^2} = \cos\theta \]

\[ \int \sin^3\theta\,\cos\theta\,\cos\theta\,d\theta = \int \sin^3\theta\cos^2\theta\,d\theta \]

\[ = \int \left(1 - \cos^2\theta\right)\cos^2\theta\,\sin\theta\,d\theta \]

\[ = \int \left(u^4 - u^2\right)du = \tfrac15u^5 - \tfrac13u^3 + C \]

\[ = \tfrac15\cos^5\theta - \tfrac13\cos^3\theta + C \]

\[ \cos\theta = \sqrt{1 - x^2}: \quad \tfrac15\left(1 - x^2\right)^{5/2} - \tfrac13\left(1 - x^2\right)^{3/2} + C \]

\[ \left[\tfrac15(1 - x^2)^{5/2} - \tfrac13(1 - x^2)^{3/2}\right]_0^1 = 0 - \left(\tfrac15 - \tfrac13\right) = \tfrac{2}{15} \]

OpenStax Calculus Volume 2, §3.3 Trigonometric Substitution §3.3, p. 289 — Example 3.23, method 2

Now the same integral by trigonometric substitution, to see that it also works and to compare the effort. The substitution turns it into sine cubed times cosine squared, an odd power of sine, which Section 3.2 handles by saving one sine and converting the others to cosines.

The substitution u equal to cosine theta then gives a polynomial in u, and you climb back through cosine theta to x using the triangle, where cosine is the root of 1 minus x squared over 1.

The result has the same two terms as method 1, just written in the other order, so the two methods agree exactly, not merely up to a constant. The definite integral from 0 to 1 is two fifteenths either way. The lesson is about efficiency: both routes are correct, and the ordinary substitution is shorter.

19. Checkpoint 3.14: set it up, stop before integrating

Step zero

\[ \int \frac{x^3}{\sqrt{25 - x^2}}\,dx \]

OpenStax Calculus Volume 2, §3.3 Trigonometric Substitution §3.3, p. 289 — Checkpoint 3.14

Write your setup before revealing. The constant under the root is 25, so a is 5 and the substitution is x equals 5 sine theta. Then x cubed becomes 125 sine cubed, the root becomes 5 cosine, and dx becomes 5 cosine d theta.

Put those in and the two factors of 5 cosine cancel, top and bottom, leaving 125 sine cubed theta. That is an odd power of sine, a Section 3.2 integral you already know how to finish.

The point of stopping here is to notice how much of the work is the setup. Once the integral is correctly written in theta, the rest is a technique you have practised. Most errors in this section happen in these first three substitutions, not in the integration.

20. The tangent substitution

Section

Part 2

21. Why x = a tan theta clears a sum of squares

Concept

\[ \sqrt{a^2 + x^2} = \sqrt{a^2 + a^2\tan^2\theta} = \sqrt{a^2\left(1 + \tan^2\theta\right)} \]

\[ = \sqrt{a^2\sec^2\theta} = |a\sec\theta| = a\sec\theta \qquad \left(-\tfrac{\pi}{2} < \theta < \tfrac{\pi}{2}\right) \]

\[ dx = a\sec^2\theta\,d\theta \]

A sum of squares has a root for every x, so the substitution must be able to produce every real number. Sine cannot; tangent can, and on the same interval secant is positive.

OpenStax Calculus Volume 2, §3.3 Trigonometric Substitution §3.3, p. 289 — Problem-Solving Strategy for a squared plus x squared

The same three moves, with a different identity. Substitute a tan theta, factor out a squared, and one plus tangent squared becomes secant squared. The root of that is the absolute value of a sec theta.

Why tangent and not sine? Look at the picture. A sum of squares has a root for every real x, so the substitution must be able to produce every real number. Sine only reaches between minus 1 and 1; tangent, on the open interval from minus a right angle to plus a right angle, reaches everything. On the same interval secant, the orange curve, stays at or above 1, so the absolute value can be dropped.

The interval is open this time because tangent is undefined at the endpoints. The book also mentions that cotangent would work, but tangent is the standard choice.

22. The tangent triangle

Concept

\[ \tan\theta = \frac{x}{a}, \quad \sec\theta = \frac{\sqrt{a^2 + x^2}}{a}, \quad \sin\theta = \frac{x}{\sqrt{a^2 + x^2}} \]

\[ \theta = \tan^{-1}\left(\frac{x}{a}\right) \]

Now the substitution names the two legs, and the root is the hypotenuse. The triangle is drawn for positive x, but its ratios are right for negative x too.

OpenStax Calculus Volume 2, §3.3 Trigonometric Substitution §3.3, p. 290 — Figure 3.7

The substitution says tangent theta is x over a, and tangent is opposite over adjacent, so the two legs are x and a. The root, the square root of a squared plus x squared, is now the hypotenuse.

From the triangle you can read every other function. Secant is hypotenuse over adjacent, which is the root over a. Sine is opposite over hypotenuse, which is x over the root. And if theta appears by itself, it is the inverse tangent of x over a.

Compare with the sine triangle: there the root was a leg, here it is the hypotenuse. In both cases it is the side Pythagoras supplies, which is why the root comes back in the answer.

23. Example 3.24: one over the root of one plus x squared

Worked example

\[ \int \frac{dx}{\sqrt{1 + x^2}} \]

\[ x = \tan\theta, \quad dx = \sec^2\theta\,d\theta, \quad \sqrt{1 + x^2} = \sec\theta \]

\[ \int \frac{\sec^2\theta}{\sec\theta}\,d\theta = \int \sec\theta\,d\theta \]

\[ = \ln|\sec\theta + \tan\theta| + C \]

\[ \sec\theta = \sqrt{1 + x^2}, \quad \tan\theta = x \]

\[ = \ln\left|\sqrt{1 + x^2} + x\right| + C \]

\[ \frac{1}{\sqrt{1 + x^2} + x}\left(\frac{x}{\sqrt{1 + x^2}} + 1\right) = \frac{1}{\sqrt{1 + x^2} + x}\cdot\frac{x + \sqrt{1 + x^2}}{\sqrt{1 + x^2}} \]

\[ = \frac{1}{\sqrt{1 + x^2}} \]

OpenStax Calculus Volume 2, §3.3 Trigonometric Substitution §3.3, pp. 290-291 — Example 3.24

With a equal to 1, the substitution is x equals tan theta. The differential brings secant squared, the root becomes secant, and one secant cancels, leaving the integral of secant alone. That is a standard antiderivative, the logarithm of secant plus tangent.

The triangle has legs x and 1, so tangent is simply x and secant is the root of 1 plus x squared. Put them in and the answer is the logarithm of the root plus x.

The book checks this one by differentiating, and you should follow it line by line. The chain rule gives one over the logarithm's argument times the derivative of the argument; putting that derivative over the common denominator makes its numerator equal to the argument, which cancels. What remains is one over the root, the integrand. The book also remarks that the argument is always positive, so the absolute value bars can be replaced by ordinary brackets.

24. A hyperbolic alternative

Concept

\[ \cosh^2\theta - \sinh^2\theta = 1 \quad\Longrightarrow\quad 1 + \sinh^2\theta = \cosh^2\theta \]

\[ x = a\sinh\theta: \quad \sqrt{a^2 + x^2} = a\cosh\theta, \quad dx = a\cosh\theta\,d\theta \]

Sinh reaches every real number and cosh is always positive, so this substitution clears the root with no interval to restrict and no absolute value to worry about.

OpenStax Calculus Volume 2, §3.3 Trigonometric Substitution §3.3, p. 291 — before Example 3.25

The hyperbolic functions have their own Pythagorean identity: cosh squared minus sinh squared equals one. Rearranged, one plus sinh squared is cosh squared, the same shape as one plus tangent squared is secant squared.

The picture shows why sinh is an even better fit than tangent for a sum of squares. Sinh, the blue curve, takes every real value, and it does so over the whole real line, with no asymptotes. Cosh, the orange curve, never drops below one, so it is always positive and the absolute value from the root disappears with no interval to worry about.

The trade-off is that you must be comfortable with hyperbolic functions and their inverses from Volume 1. The next example redoes Example 3.24 this way.

25. Example 3.25: the same integral with sinh

Worked example

\[ \int \frac{dx}{\sqrt{1 + x^2}} \]

\[ x = \sinh\theta, \quad dx = \cosh\theta\,d\theta \]

\[ \sqrt{1 + \sinh^2\theta} = \sqrt{\cosh^2\theta} = \cosh\theta \]

\[ \int \frac{\cosh\theta}{\cosh\theta}\,d\theta = \int 1\,d\theta \]

\[ = \theta + C = \sinh^{-1}x + C \]

\[ \frac{d}{dx}\sinh^{-1}x = \frac{1}{\sqrt{1 + x^2}} \]

OpenStax Calculus Volume 2, §3.3 Trigonometric Substitution §3.3, p. 291 — Example 3.25

Watch how short this is. With x equal to sinh theta, dx is cosh theta d theta, and the root of 1 plus sinh squared is cosh. The integrand collapses to one, so the integral is just theta, which is the inverse hyperbolic sine of x.

The simplification from the root of cosh squared to cosh is safe for every theta, because cosh is always positive. There is no range of theta to track.

The check uses the derivative of the inverse hyperbolic sine, one over the root of 1 plus x squared, which you met in Volume 1. So the answer is right. But it looks nothing like the answer from Example 3.24, and the next slide settles whether two different-looking answers can both be correct.

26. Example 3.25, analysis: two answers, one function

Worked example

Example 3.24 gave a logarithm; Example 3.25 gave an inverse hyperbolic sine. Show they are the same function.

\[ y = \sinh^{-1}x \iff \sinh y = x \iff \frac{e^y - e^{-y}}{2} = x \]

\[ e^{2y} - 1 = 2xe^y \iff e^{2y} - 2xe^y - 1 = 0 \]

\[ e^y = \frac{2x \pm \sqrt{4x^2 + 4}}{2} = x \pm \sqrt{x^2 + 1} \]

\[ x - \sqrt{x^2 + 1} < 0 \;\Longrightarrow\; e^y = x + \sqrt{x^2 + 1} \]

\[ \sinh^{-1}x = \ln\left(x + \sqrt{x^2 + 1}\right) \]

\[ \sinh^{-1}1 = 0.881374, \qquad \ln\left(1 + \sqrt2\right) = 0.881374 \]

OpenStax Calculus Volume 2, §3.3 Trigonometric Substitution §3.3, pp. 291-292 — Example 3.25, Analysis

Two correct antiderivatives of the same function can differ only by a constant, so either these two answers are the same function or one of them is wrong. The way to find out is to solve for the inverse hyperbolic sine explicitly.

Write y for it, so sinh y equals x. Writing sinh with exponentials and clearing the fractions gives a quadratic equation, not in y but in e to the y. The quadratic formula gives two candidates, x plus the root and x minus the root. The second is always negative, because the root of x squared plus 1 is bigger than x, and an exponential cannot be negative, so it is discarded.

Taking logarithms shows the inverse hyperbolic sine of x is exactly the logarithm of x plus the root: the two answers are identical, with no constant between them. The numerical check at x equal to 1 confirms it to six places.

27. Limits travel with the substitution

Picture it

For a definite integral you need not go back to x at all. Push each limit through the substitution once, at the start, and finish in theta.

\[ x = \tfrac12\tan\theta: \quad x = 0 \to \theta = 0, \qquad x = \tfrac12 \to \theta = \tfrac{\pi}{4} \]

The curve is the substitution itself, x as a function of theta. Follow the dashed guides: theta equal to zero gives x equal to zero, and theta equal to pi over 4 gives x equal to one half, because tangent of pi over 4 is 1.

So the stretch of x from 0 to one half, marked on the vertical axis, is the same as the stretch of theta from 0 to pi over 4, marked on the horizontal axis. A definite integral over one is the same number as the converted integral over the other.

This saves real work. Once the limits are in theta, you evaluate the theta antiderivative at theta values and never need a reference triangle. The next two slides use it on an arc-length problem.

28. How long is the parabola from 0 to one half?

Estimation

Make a guess before revealing, and use the picture to make it an informed one. The arc runs from the origin to the point one half, one quarter, so it cannot be shorter than the straight chord between them.

The chord has length the root of one quarter plus one sixteenth, about 0.559. The arc bends only slightly above it, so its length should be just a little more. That rules out anything under one half, which would be shorter than the horizontal run alone, and it makes 0.75 or 1 far too long.

The exact answer, which the next slide computes, is about 0.5739. Having a rough expected value before a long calculation is the best way to catch a slip at the end of it.

29. Example 3.26: the length of a parabola

Worked example

\[ L = \int_0^{1/2}\sqrt{1 + (2x)^2}\,dx = \int_0^{1/2}\sqrt{1 + 4x^2}\,dx \]

\[ x = \tfrac12\tan\theta, \quad dx = \tfrac12\sec^2\theta\,d\theta \]

\[ x = 0 \to \theta = 0, \qquad x = \tfrac12 \to \tan\theta = 1 \to \theta = \tfrac{\pi}{4} \]

\[ L = \int_0^{\pi/4}\sec\theta\cdot\tfrac12\sec^2\theta\,d\theta = \tfrac12\int_0^{\pi/4}\sec^3\theta\,d\theta \]

\[ = \tfrac14\left[\sec\theta\tan\theta + \ln|\sec\theta + \tan\theta|\right]_0^{\pi/4} \]

\[ = \tfrac14\left[\sqrt2 + \ln\left(\sqrt2 + 1\right)\right] - \tfrac14\left[0 + \ln 1\right] \]

\[ L = \tfrac14\left(\sqrt2 + \ln\left(\sqrt2 + 1\right)\right) \approx 0.57390 \]

\[ \text{Simpson } (n = 2000): 0.57390, \qquad \text{chord } = 0.55902 < 0.57390 \]

OpenStax Calculus Volume 2, §3.3 Trigonometric Substitution §3.3, pp. 292-293 — Example 3.26

Arc length needs the root of one plus the derivative squared, and for y equal to x squared that is the root of 1 plus 4 x squared, a sum of squares. Seeing 4 x squared as 2x squared tells you to set 2x equal to tan theta, so x is one half tan theta.

Convert the limits immediately: x equal to one half means tangent theta is 1, so theta is pi over 4. The root becomes secant, the differential brings one half secant squared, and the integral is one half the integral of secant cubed, the result derived by parts in Section 3.2.

Evaluating at pi over 4, where secant is the root of 2 and tangent is 1, gives one quarter of the root of 2 plus the logarithm of the root of 2 plus 1, about 0.5739. Simpson's rule on the original integrand agrees, and the answer is a little more than the chord, just as the estimate predicted. The figure shows the same number as an area under the arc-length integrand.

30. Checkpoint 3.15: rewrite with a tangent

Fill the middle

Substitute x equal to 2 tan theta, clear the root, and collect every constant into one number.

\[ \int x^3\sqrt{x^2 + 4}\,dx = \int C\,\tan^3\theta\,\sec^3\theta\,d\theta \]

OpenStax Calculus Volume 2, §3.3 Trigonometric Substitution §3.3, p. 293 — Checkpoint 3.15

Fill in the constant before checking. The root of x squared plus 4 is a sum of squares with a equal to 2, so x is 2 tan theta.

Now convert each piece. x cubed is 8 tan cubed theta. The root is 2 sec theta. The differential is 2 sec squared theta d theta. Multiply: the constants give 8 times 2 times 2, which is 32, the tangents give tan cubed, and the secants, one from the root and two from the differential, give sec cubed.

The checkpoint asks only for this rewriting, and that is a good habit to practise on its own: getting all three conversions right, and collecting the constants carefully, is where most of the marks are.

31. Exercise 137: a coefficient on x squared

Worked example

\[ \int \frac{dx}{\sqrt{1 + 9x^2}} \]

\[ 1 + 9x^2 = 1 + (3x)^2 \;\Longrightarrow\; 3x = \tan\theta \]

\[ x = \tfrac13\tan\theta, \quad dx = \tfrac13\sec^2\theta\,d\theta, \quad \sqrt{1 + 9x^2} = \sec\theta \]

\[ \int \frac{\tfrac13\sec^2\theta}{\sec\theta}\,d\theta = \tfrac13\int \sec\theta\,d\theta \]

\[ = \tfrac13\ln|\sec\theta + \tan\theta| + C \]

\[ \tan\theta = 3x, \quad \sec\theta = \sqrt{1 + 9x^2} \]

\[ = \tfrac13\ln\left|\sqrt{1 + 9x^2} + 3x\right| + C \]

\[ \frac13\cdot\frac{\frac{9x}{\sqrt{1 + 9x^2}} + 3}{\sqrt{1 + 9x^2} + 3x} = \frac13\cdot\frac{3}{\sqrt{1 + 9x^2}} = \frac{1}{\sqrt{1 + 9x^2}} \]

OpenStax Calculus Volume 2, §3.3 Trigonometric Substitution §3.3, p. 296 — Exercise 137

A coefficient on x squared does not change which substitution to use, only how you write it. Nine x squared is 3x squared, so this is one plus a square, and 3x plays the role of x: set 3x equal to tan theta.

Solving for x gives one third tan theta, and the differential inherits the one third. The root becomes secant, one secant cancels, and you are left with one third of the integral of secant.

For the triangle, the opposite side is 3x and the adjacent is 1, so tangent is 3x and secant is the root of 1 plus 9 x squared. The check by differentiating is worth doing slowly: the derivative of the logarithm's argument is 9x over the root plus 3, which is 3 times the argument divided by the root, so the argument cancels and the one third meets the 3.

32. Find the error: the old limits kept

Error analysis

Read the line and find the problem before revealing the notes. The substitution itself was done correctly; the integrand in theta is right.

The mistake is in the limits. Zero and one half describe x, but the new integral runs over theta, so its limits must be the theta values that correspond: zero and pi over 4. Keeping the old limits computes the area under the new integrand over the wrong interval, and gives about 0.286 instead of 0.574.

There is a built-in alarm here: 0.286 is shorter than the straight chord between the endpoints, and no curve joining two points can be shorter than the segment joining them. You can fix it either way the note says, converting the limits or converting back to x first, but never by mixing the two.

33. The secant substitution

Section

Part 3

34. Why x = a sec theta clears the other difference

Concept

\[ \sqrt{x^2 - a^2} = \sqrt{a^2\sec^2\theta - a^2} = \sqrt{a^2\left(\sec^2\theta - 1\right)} \]

\[ = \sqrt{a^2\tan^2\theta} = |a\tan\theta|, \qquad dx = a\sec\theta\tan\theta\,d\theta \]

This root exists only when x is at least a or at most minus a, the two branches of secant. On the right branch tangent is negative, so the absolute value has to stay until you know which branch you are on.

OpenStax Calculus Volume 2, §3.3 Trigonometric Substitution §3.3, p. 293 — Problem-Solving Strategy for x squared minus a squared

The third form is a square minus a constant, the root of x squared minus a squared. The identity with that shape is secant squared minus one equals tangent squared, so substitute x equals a sec theta. The root becomes the absolute value of a tan theta.

This root only exists for x at least a or at most minus a, and those two pieces match the two branches of secant in the picture. On the left branch, theta between zero and a right angle, tangent is positive. On the right branch, theta between a right angle and pi, secant is negative and so is tangent, the dashed curve.

That is why this form, unlike the other two, keeps its absolute value. You drop it only after you know which branch x is on, which usually the problem tells you.

35. Reading the secant substitution

Notation

OpenStax Calculus Volume 2, §3.3 Trigonometric Substitution §3.3, p. 293 — Problem-Solving Strategy, step 2

Go through the notes one at a time. The interval for theta has two pieces, one for each branch of secant, and the right angle between them is excluded because secant is undefined there.

The differential is where secant substitutions get their tidiness. The derivative of secant is secant times tangent, so dx carries a factor of tangent, and it very often cancels the tangent that the root produced. You will see that in Checkpoint 3.16.

The third note is the one to remember: the root is a tan theta on the first branch and minus a tan theta on the second. The book therefore gives two reference triangles for this form, one for each sign of x.

36. Predict the root on the left branch

Prediction

\[ x = 3\sec\theta, \qquad x = -5 \]

Commit to an answer first. The tempting reply is 4, because the root of 25 minus 9 is 4. But the question asks for 3 tan theta, and that is a different thing on this branch.

With x equal to minus 5, secant theta is minus five thirds, so cosine theta is minus three fifths and theta lies past a right angle. There sine is positive, four fifths, and tangent is sine over cosine, minus four thirds. So 3 tan theta is minus 4.

Meanwhile the root of x squared minus 9 is plus 4, as every root is. The two differ by a sign, which is exactly the absolute value in the substitution: on the left branch the root equals minus 3 tan theta.

37. Two triangles for the secant

Picture it

For x at least a, the familiar triangle: hypotenuse x, adjacent a, opposite the root. For x at most minus a, theta is obtuse and the horizontal leg lies on the negative side, so every ratio that uses it changes sign.

OpenStax Calculus Volume 2, §3.3 Trigonometric Substitution §3.3, p. 294 — Figure 3.9

The left triangle is the one to use when x is at least a. Secant is hypotenuse over adjacent, so the hypotenuse is x and the adjacent side is a, and Pythagoras gives the opposite side as the root of x squared minus a squared. Tangent is that root over a, positive.

The right picture handles x at most minus a. Now theta is obtuse, drawn in standard position with its terminal side up and to the left. The horizontal leg points in the negative direction, so it is labelled minus a, and the slanted side has length the absolute value of x, which is minus x because x is negative.

Read tangent as vertical over horizontal, the root over minus a, and it comes out negative. The root itself, a length, is still positive. That single sign is the whole difference between the two cases.

38. Trap: dropping the absolute value on the left branch

Trap

39. Example 3.27: the area under a hyperbola

Worked example

\[ A = \int_3^5 \sqrt{x^2 - 9}\,dx \]

\[ x = 3\sec\theta, \quad dx = 3\sec\theta\tan\theta\,d\theta, \quad \sqrt{x^2 - 9} = 3\tan\theta \]

\[ x = 3 \to \theta = 0, \qquad x = 5 \to \theta = \sec^{-1}\tfrac53 \]

\[ A = \int_0^{\sec^{-1}(5/3)} 9\tan^2\theta\sec\theta\,d\theta \]

\[ = \int_0^{\sec^{-1}(5/3)} 9\left(\sec^3\theta - \sec\theta\right)d\theta \]

\[ = \left[\tfrac92\sec\theta\tan\theta - \tfrac92\ln|\sec\theta + \tan\theta|\right]_0^{\sec^{-1}(5/3)} \]

\[ = \tfrac92\cdot\tfrac53\cdot\tfrac43 - \tfrac92\ln\left(\tfrac53 + \tfrac43\right) - \left(0 - \tfrac92\ln 1\right) \]

\[ A = 10 - \tfrac92\ln 3 \approx 10 - 4.94376 = 5.05624 \]

\[ \int_3^5\sqrt{x^2 - 9}\,dx \approx 5.05624 \quad (\text{Simpson}) \]

OpenStax Calculus Volume 2, §3.3 Trigonometric Substitution §3.3, pp. 294-295 — Example 3.27

The region runs from x equal to 3 to 5, on the right branch, so the root is 3 tan theta with no sign trouble. Convert the limits: x equal to 3 means secant is 1 and theta is zero, and x equal to 5 means theta is the inverse secant of five thirds.

The integrand becomes 9 tan squared times secant. Replace tan squared by secant squared minus one, and you have two integrals you know: secant cubed and secant. Their logarithm terms combine, nine halves of one minus nine of the other, leaving minus nine halves of a single logarithm.

At the upper limit, secant is five thirds and, from the triangle with hypotenuse 5 and adjacent 3, tangent is four thirds. The product term gives 10 and the logarithm is of five thirds plus four thirds, which is 3. So the area is 10 minus nine halves ln 3, about 5.056, and Simpson's rule agrees. The figure shows a region a bit wider than 2 and up to 4 high, so an area near 5 is believable.

40. Find the error: a sign in the identity

Error analysis

Try to spot the problem before revealing. The substitution into the integrand is fine; the trouble is the identity used to rewrite tan squared.

The correct identity comes from dividing sine squared plus cosine squared equals one by cosine squared: tangent squared plus one equals secant squared. So tangent squared is secant squared minus one. The version on the slide has the subtraction reversed, and a one-second test exposes it: the left side is a square and can never be negative, while one minus secant squared can never be positive.

It is worth knowing that this exact slip appears in the textbook's margin note for Example 3.27, while the working beside it uses the correct identity. Printed sources make mistakes too, and the sign test is how you catch them.

41. Checkpoint 3.16: one over the root of x squared minus 4

Worked example

\[ \int \frac{dx}{\sqrt{x^2 - 4}}, \qquad x > 2 \]

\[ x = 2\sec\theta, \quad dx = 2\sec\theta\tan\theta\,d\theta, \quad \sqrt{x^2 - 4} = 2\tan\theta \]

\[ \int \frac{2\sec\theta\tan\theta}{2\tan\theta}\,d\theta = \int \sec\theta\,d\theta \]

\[ = \ln|\sec\theta + \tan\theta| + C \]

\[ \sec\theta = \tfrac{x}{2}, \quad \tan\theta = \tfrac{\sqrt{x^2 - 4}}{2} \]

\[ = \ln\left|\frac{x}{2} + \frac{\sqrt{x^2 - 4}}{2}\right| + C \]

\[ \frac{d}{dx}\ln\left(x + \sqrt{x^2 - 4}\right) = \frac{1 + \frac{x}{\sqrt{x^2 - 4}}}{x + \sqrt{x^2 - 4}} = \frac{1}{\sqrt{x^2 - 4}} \]

OpenStax Calculus Volume 2, §3.3 Trigonometric Substitution §3.3, p. 295 — Checkpoint 3.16

The problem says x is greater than 2, which puts you on the right branch, so the root is simply 2 tan theta. This is the reason the book adds that condition.

Now the secant substitution shows its best feature: the differential is 2 sec theta tan theta d theta, and its tangent cancels the tangent from the root. What is left is the integral of secant, the logarithm of secant plus tangent.

The triangle, hypotenuse x and adjacent 2, gives secant as x over 2 and tangent as the root over 2. For the check, notice the logarithm of a sum over 2 is the logarithm of the sum minus the logarithm of 2, a constant, so you can differentiate the simpler logarithm of x plus the root. Its derivative collapses to one over the root, just as in Example 3.24.

42. Choosing the substitution

Section

Part 4

43. One identity per form

Intuition

Read the radical, find the identity with the same shape, and the substitution follows. Which square comes first matters: the constant first means sine, the variable first means secant.

OpenStax Calculus Volume 2, §3.3 Trigonometric Substitution §3.3, pp. 286-293 — the three problem-solving strategies

The table is the whole choice in three rows, and it is better to understand it than to memorise it. Each radical is a sum or difference of two squares, and each identity is a sum or difference of two squares that equals a single square. Match the shapes and the substitution names itself.

The detail that trips people is the order in a difference. A constant minus the variable squared matches one minus sine squared, so sine. The variable squared minus a constant matches secant squared minus one, so secant. Swap them and the identity no longer produces a single square.

If you ever forget a row, rebuild it: ask which trigonometric function, squared, you can add one to or subtract from one to get another square.

44. Sine, tangent or secant?

Sorting

Sort each radical before checking. Look only at the pattern: is it a constant minus a square, a sum of two squares, or a square minus a constant?

Three of these have a coefficient on x squared, and the coefficient changes nothing about the choice. Nine x squared is 3x squared, four x squared is 2x squared, so the square is of 3x or 2x rather than of x, and the constant a is whatever is left. For the root of 1 minus 9 x squared you set 3x equal to sine theta; for the root of 4 x squared plus 9 you set 2x equal to 3 tan theta.

If you misplaced the root of 9 x squared minus 1, check the order: the variable term comes first, so it is the secant form.

45. Completing the square reaches a standard form

Concept

\[ -x^2 + 2x + 8 = -\left(x^2 - 2x\right) + 8 \]

\[ = -\left[(x - 1)^2 - 1\right] + 8 = 9 - (x - 1)^2 \]

With u equal to x minus 1, this is the sine form with a equal to 3. Every quadratic under a root, once the square is completed, is a sum or difference of two squares, so the three forms cover them all.

OpenStax Calculus Volume 2, §3.3 Trigonometric Substitution §3.3, pp. 296-297 — Exercises 131-133 and 160-164

Real integrals do not always arrive with a bare x squared. A quadratic with an x term, like minus x squared plus 2x plus 8, has to be rewritten first, and completing the square does it.

Factor the minus sign out of the x terms, complete the square inside the bracket by adding and subtracting 1, and bring the minus 1 out as plus 1. The result is 9 minus x minus 1 squared: the sine form, in the shifted variable x minus 1.

The pictures show what happened. On the left, completing the square found the vertex of the parabola at x equal to 1, height 9. On the right, the root of the same quadratic is a semicircle of radius 3 centred at x equal to 1. Every quadratic under a root is a shifted and scaled version of one of the three forms, which is why the list of three is complete.

46. Exercise 162: a quadratic under the root

Worked example

\[ \int \frac{dx}{\sqrt{-x^2 + 2x + 8}} \]

\[ -x^2 + 2x + 8 = 9 - (x - 1)^2 \]

\[ x - 1 = 3\sin\theta, \quad dx = 3\cos\theta\,d\theta, \quad \sqrt{9 - (x - 1)^2} = 3\cos\theta \]

\[ \int \frac{3\cos\theta}{3\cos\theta}\,d\theta = \int d\theta = \theta + C \]

\[ = \sin^{-1}\left(\frac{x - 1}{3}\right) + C \]

\[ \frac{1}{\sqrt{1 - \frac{(x - 1)^2}{9}}}\cdot\frac13 = \frac{1}{\sqrt{9 - (x - 1)^2}} = \frac{1}{\sqrt{-x^2 + 2x + 8}} \]

OpenStax Calculus Volume 2, §3.3 Trigonometric Substitution §3.3, p. 297 — Exercise 162

With the square completed on the previous slide, the integrand is one over the root of 9 minus x minus 1 squared. The substitution works on the shifted variable: x minus 1 equals 3 sine theta, and dx is still 3 cosine theta d theta because the shift has derivative zero.

Everything cancels, leaving the integral of d theta, which is theta. Undo the substitution: theta is the inverse sine of x minus 1 over 3.

The check differentiates the inverse sine with the chain rule. The factor of one third from the inside combines with the fraction under the root to give one over the root of 9 minus x minus 1 squared, and expanding that recovers the original quadratic. You could also have recognised this directly as an inverse-sine formula after completing the square; either way, completing the square is the key step.

47. Exercise 160: completing the square into the secant form

Worked example

\[ \int \frac{dx}{\sqrt{x^2 - 6x}}, \qquad x > 6 \]

\[ x^2 - 6x = (x - 3)^2 - 9 \]

\[ x - 3 = 3\sec\theta, \quad dx = 3\sec\theta\tan\theta\,d\theta, \quad \sqrt{(x - 3)^2 - 9} = 3\tan\theta \]

\[ \int \frac{3\sec\theta\tan\theta}{3\tan\theta}\,d\theta = \int \sec\theta\,d\theta = \ln|\sec\theta + \tan\theta| + C \]

\[ \sec\theta = \frac{x - 3}{3}, \quad \tan\theta = \frac{\sqrt{x^2 - 6x}}{3} \]

\[ F(x) = \ln\left|\frac{x - 3 + \sqrt{x^2 - 6x}}{3}\right| + C \]

\[ F(9) - F(7) = 0.52159, \qquad \int_7^9 \frac{dx}{\sqrt{x^2 - 6x}} \approx 0.52159 \]

OpenStax Calculus Volume 2, §3.3 Trigonometric Substitution §3.3, p. 297 — Exercise 160

Half of minus 6 is minus 3, so x squared minus 6x is x minus 3 squared minus 9: a square minus a constant, the secant form, with a equal to 3. The root only exists for x at most 0 or at least 6, and the condition x greater than 6 puts you on the right branch.

Set x minus 3 equal to 3 sec theta. The tangent in the differential cancels the tangent from the root, exactly as in Checkpoint 3.16, and the integral is the logarithm of secant plus tangent.

The triangle now has hypotenuse x minus 3 and adjacent side 3, so secant is x minus 3 over 3 and tangent is the root of x squared minus 6x, over 3. The check compares the change in the antiderivative from 7 to 9 with Simpson's rule on the original integrand; they agree to five places.

48. Match each quadratic to its completed square

Matching

Match each quadratic before checking. Complete the square yourself, then decide the form from what is left over.

For x squared plus 4x plus 13, half of 4 is 2, and 13 minus 4 leaves 9 added: a sum of squares, tangent. For x squared plus 4x minus 12, the same shift leaves minus 16: a square minus a constant, secant. The two with a leading minus need the sign factored out first: minus x squared plus 10x becomes 25 minus x minus 5 squared, and minus x squared minus 2x plus 4 becomes 5 minus x plus 1 squared, both sine forms.

Expanding is always a quick check. If your completed square does not expand back to the original, find the error before you substitute anything.

49. Trap: completing the square with a leading minus

Trap

50. Definite integrals, and when not to substitute

Section

Part 5

51. Convert the limits once, never convert back

Concept

\[ \int_{x_0}^{x_1} f(x)\,dx = \int_{\theta_0}^{\theta_1} f(g(\theta))\,g'(\theta)\,d\theta, \qquad x_0 = g(\theta_0),\; x_1 = g(\theta_1) \]

The substitution x equal to g of theta carries the endpoints with it. Converting the limits replaces the whole reference-triangle step: evaluate in theta and you are done.

The triangle is still needed for an indefinite integral, where there is nothing to evaluate and the answer must be a function of x.

OpenStax Calculus Volume 2, §3.3 Trigonometric Substitution §3.3, pp. 292-295 — Examples 3.26 and 3.27

This is the general rule behind Example 3.26 and Example 3.27. A substitution x equals g of theta changes the integrand, the differential and the limits together, and once all three have changed you have an integral purely in theta, which you evaluate at theta values.

That means no reference triangle and no inverse trigonometric functions of x at the end, which is often the longest part of an indefinite problem. The trade is that you must find the theta limits, which means solving the substitution for theta at each endpoint.

For an indefinite integral there are no limits to convert, and the answer must be a function of x, so the triangle is unavoidable there. Choose the finish that matches the question.

52. Exercise 165: half a disc, by calculus

Worked example

\[ \int_{-3}^{3}\sqrt{9 - x^2}\,dx \]

\[ x = 3\sin\theta: \quad x = -3 \to \theta = -\tfrac{\pi}{2}, \quad x = 3 \to \theta = \tfrac{\pi}{2} \]

\[ \int_{-\pi/2}^{\pi/2} 9\cos^2\theta\,d\theta \]

\[ = \left[\tfrac92\theta + \tfrac94\sin 2\theta\right]_{-\pi/2}^{\pi/2} \]

\[ = \tfrac92\cdot\tfrac{\pi}{2} - \tfrac92\cdot\left(-\tfrac{\pi}{2}\right) = \tfrac{9\pi}{2} \approx 14.137 \]

\[ \tfrac12\pi r^2 = \tfrac12\pi(3)^2 = \tfrac{9\pi}{2} \]

OpenStax Calculus Volume 2, §3.3 Trigonometric Substitution §3.3, p. 297 — Exercise 165

The exercise asks you to find this without calculus, and you should: it is the area of half a disc of radius 3. But it is also a clean example of converted limits, because the endpoints minus 3 and 3 map to theta equal to minus and plus pi over 2.

The figure shows the substitution at work. Each ray is a value of theta, measured from the vertical, and it ends at the point whose horizontal coordinate is 3 sine theta. As theta sweeps from minus pi over 2 to pi over 2, that point sweeps x across the whole diameter exactly once.

The integrand becomes 9 cosine squared, the half-angle identity integrates it, and the sine of 2 theta term vanishes at both limits. What is left is 9 pi over 2, precisely half of pi times 3 squared. Calculus and geometry agree, as they must.

53. Exercise 166: the area of an ellipse

Worked example

\[ \frac{x^2}{4} + \frac{y^2}{9} = 1 \]

\[ y = 3\sqrt{1 - \tfrac{x^2}{4}} = \tfrac32\sqrt{4 - x^2} \]

\[ A = 4\int_0^2 \tfrac32\sqrt{4 - x^2}\,dx = 6\int_0^2\sqrt{4 - x^2}\,dx \]

\[ \sqrt{4 - x^2}\,dx = 2\cos\theta\cdot 2\cos\theta\,d\theta = 4\cos^2\theta\,d\theta \]

\[ A = 24\int_0^{\pi/2}\cos^2\theta\,d\theta = 24\left[\tfrac{\theta}{2} + \tfrac{\sin 2\theta}{4}\right]_0^{\pi/2} \]

\[ A = 24\cdot\tfrac{\pi}{4} = 6\pi \approx 18.850 \]

\[ \pi ab = \pi(2)(3) = 6\pi \]

OpenStax Calculus Volume 2, §3.3 Trigonometric Substitution §3.3, p. 297 — Exercise 166

Solve the ellipse for y and take the top half: y is three halves the root of 4 minus x squared. The ellipse is symmetric about both axes, so its area is four times the shaded quarter, which gives 6 times the integral of the root from 0 to 2.

That integral is the sine form with a equal to 2. With the limits converted to 0 and pi over 2, the root and the differential together give 4 cosine squared, so the area is 24 times the integral of cosine squared over a quarter turn, which is pi over 4.

The answer, 6 pi, matches the well-known formula pi times the two semi-axes, here 2 and 3. And if both semi-axes were equal to r, the same calculation would give pi r squared, the area of a circle, which is exactly what an ellipse with equal axes is.

54. When an ordinary substitution is quicker

Concept

Step 1 of every strategy in the book: check for an easier method first. A spare factor of x outside the root is the usual signal.

OpenStax Calculus Volume 2, §3.3 Trigonometric Substitution §3.3, p. 286 — Problem-Solving Strategy, step 1; Exercise 170

Every strategy box in the book opens with the same step: check whether the integral can be done more easily another way. Trigonometric substitution always works on these forms, but it is the long way round when a short one exists.

The signal to look for is a spare odd power of x outside the root. Compare the first two rows of the table: with an x outside, u equal to the radicand makes du absorb it and the integral is one line; without it, u leaves an x behind with nowhere to go, and the tangent substitution is needed.

The last row is a reminder that some roots are already standard formulas. One over the root of 4 minus x squared is an inverse sine from Volume 1, and recognising that saves the whole substitution.

55. Exercise 169: one integral, two routes

Worked example

\[ \int \frac{x}{\sqrt{x^2 + 1}}\,dx \]

\[ u = x^2 + 1, \quad du = 2x\,dx: \quad \tfrac12\int u^{-1/2}\,du = \sqrt{u} = \sqrt{x^2 + 1} + C \]

\[ dx = \sec^2\theta\,d\theta, \quad \sqrt{x^2 + 1} = \sec\theta \]

\[ \int \frac{\tan\theta}{\sec\theta}\,\sec^2\theta\,d\theta = \int \sec\theta\tan\theta\,d\theta \]

\[ = \sec\theta + C = \sqrt{x^2 + 1} + C \]

\[ \frac{d}{dx}\sqrt{x^2 + 1} = \frac{2x}{2\sqrt{x^2 + 1}} = \frac{x}{\sqrt{x^2 + 1}} \]

OpenStax Calculus Volume 2, §3.3 Trigonometric Substitution §3.3, p. 297 — Exercise 169

This exercise asks you to do the same integral both ways, and it is worth actually doing, because it shows the two methods meeting. Route 1 is one line: u is x squared plus 1, x dx is half of du, and the integral of u to the minus one half is twice the root.

Route 2 takes a few more lines. With x equal to tan theta, the integrand becomes tangent over secant times secant squared, which simplifies to secant times tangent, whose antiderivative is secant. The triangle turns secant back into the root of x squared plus 1.

Both routes give the same function, and the derivative confirms it. When two methods apply, the one that sees the spare x is almost always shorter.

56. Trigonometric substitution, or something quicker?

Discrimination

Sort each integral by the quickest method that works, not by every method that works. The test is simple: is there an odd power of x outside the root?

Three items have one, and each yields to u equal to the radicand. The one with x cubed is Example 3.23: the trigonometric route also worked there, but it took twice as long. The other three have nothing outside, or an even power, so u would leave a stray x; they need the substitution their form dictates.

If you put x squared over the root of x squared minus 1 in the quick bucket, try it: with u equal to x squared minus 1, du is 2x dx, and you would need x squared divided by x, a leftover x that cannot be written nicely in u. That leftover is the sign that a trigonometric substitution is needed.

57. The electric field of a long charged wire

Real world

OpenStax Calculus Volume 2, §3.3 Trigonometric Substitution §3.3, p. 296 — Exercise 146, the same integral with d equal to 1

Physics produces exactly these integrals. Each small piece of a charged rod pushes on a test charge at P, and adding up the parts of those pushes that point straight away from the rod gives the integral of d over x squared plus d squared to the power three halves. That power is a root of a sum of squares, cubed.

The tangent substitution, x equal to d tan theta, is not just algebra here: theta is literally the angle at P between the vertical and the line to the piece of rod, as the figure shows. The integrand collapses to cosine over d squared, whose antiderivative is sine over d squared, and the triangle converts it back.

For an infinitely long rod, x over the root of x squared plus d squared tends to 1 at one end and minus 1 at the other, so the integral is 2 over d. The field falls off like one over the distance, much more slowly than the one over distance squared of a point charge. Setting d equal to 1 gives Exercise 146 of the book, and a numerical check over a very long rod gives 1.99999.

58. Putting it together

Section

Part 6

59. Pattern: integrating a root of a quadratic

Pattern

1. Look for a quicker route first: a spare x outside the root means u = the radicand; a bare 1 over a root may be a standard inverse-trig formula.
2. Complete the square if the quadratic has an x term, and factor out any leading minus sign first.
3. Let the form choose: a² − x² takes sine, a² + x² takes tangent, x² − a² takes secant. Substitute for x, for dx, and for the root.
4. Integrate in θ with the Section 3.2 techniques.
5. Finish: for a definite integral, convert the limits; for an indefinite one, draw the triangle and write every function of θ in x.

This is the order to work in. It runs from the cheapest check to the most work, because an earlier step can end the problem.

First look for a spare odd power of x outside the root, or a standard formula; either can save the whole substitution. Then, if the quadratic has an x term, complete the square, factoring out any leading minus sign first. Now the form of the radical names the substitution, and you substitute three things: x, dx and the root.

After integrating in theta, finish the way the question requires. A definite integral: convert the limits and evaluate in theta. An indefinite one: draw the triangle for that form and rewrite every function of theta in x. For the secant form, decide the branch before dropping the absolute value.

60. The three forms side by side

Comparison

Fill in every blank before checking. Each row should be derivable from the one identity, so if a blank stumps you, rebuild it from the identity rather than trying to remember it.

The column that people most often get wrong is the last one. The sine interval is closed, because sine is fine at a right angle. The tangent interval is open, because tangent blows up there. The secant interval has two pieces with the right angle cut out, and on the second piece the tangent is negative, which is why the secant row carries an absolute value that the other two do not.

61. Put the strategy in order

Ranking

Put the steps in order before checking. This is the book's five-step problem-solving strategy for the sine form; the tangent and secant versions have the same five steps with a different substitution.

The order has reasons. The check for an easier method comes first because it might make every other step unnecessary. Substituting and simplifying the root come before integrating because the integral cannot be done until the root is gone. The triangle comes last because it converts an answer, and there is no answer to convert until you have integrated.

62. Check: choosing the substitution

Check

Write the radicand as two squares before choosing. Four x squared is 2x squared and 9 is 3 squared, so this is a sum of squares, the tangent form, with 2x playing the role of x and 3 playing the role of a.

Setting 2x equal to 3 tan theta, that is x equal to three halves tan theta, makes four x squared equal to 9 tan squared, and the radicand becomes 9 secant squared, a perfect square. The other tangent options scale x by the wrong amount, so the two terms no longer share the factor 9 and the identity cannot combine them.

63. Check: reading the triangle

Check

Draw the triangle before choosing. Sine theta is x over 3, so the side opposite theta is x and the hypotenuse is 3. Pythagoras gives the adjacent side as the root of 9 minus x squared.

Tangent is opposite over adjacent: x over the root. Each wrong option is another ratio from the same triangle, the cotangent, the sine and the secant. If you picked one of those, the triangle was right and the definition of tangent slipped, which is worth fixing now because every conversion in this section depends on it.

64. Check: converting the limits

Check

Solve the substitution for theta at each endpoint. At x equal to 2, the equation 2 tan theta equals 2 gives tan theta equal to 1, so theta is pi over 4. At x equal to 0, theta is 0.

The most tempting wrong answer is the inverse tangent of 2, which comes from setting tan theta equal to x and forgetting the factor 2 in the substitution. The other trap is keeping the x-limits unchanged, the error from the earlier error-analysis slide. Always substitute the endpoint into the substitution itself, not into a remembered shortcut.

65. Explain the third side

Explain it to yourself

Write your two sentences before revealing. This is the idea that ties together every conversion in the lesson.

The substitution tells you one trigonometric ratio as x over a or a over x, and that fixes two sides of a right triangle. The third side is determined by Pythagoras, as the square root of the sum or difference of the other two squared. And the substitution was chosen so that this sum or difference is precisely the expression under the root in the integrand. So the missing side is the original radical, every time, which is why it reappears in the answer.

66. Exit ticket

Exit ticket

\[ \int \frac{\sqrt{x^2 - 25}}{x}\,dx, \qquad x > 5 \]

Do the whole problem before revealing: name the form, substitute, integrate, convert back.

The root of x squared minus 25 is the secant form with a equal to 5, and x greater than 5 puts you on the right branch. The root becomes 5 tan theta, the differential 5 sec theta tan theta, and the x in the denominator 5 sec theta, so the secants cancel and you are left with 5 tan squared theta. Replace tan squared by secant squared minus one and integrate to 5 tan theta minus 5 theta.

The triangle, hypotenuse x and adjacent 5, makes 5 tan theta the root of x squared minus 25, and theta is the inverse secant of x over 5. The numerical check, comparing the change from 6 to 10 with Simpson's rule, confirms the answer.

67. Recap

Recap

Complete the square to reach a form, check for a quicker substitution first, convert the limits of a definite integral, and keep the absolute value on the secant's left branch.

Next, Section 3.4 handles rational functions by partial fractions, and some of its integrals end in an inverse tangent.

Stewart, Calculus: Early Transcendentals 8e, §7.3 Trigonometric Substitution §7.3, pp. 486-492 — the same material in Stewart

Three forms, three identities, three substitutions, three triangles. A constant minus a square takes sine; a sum of squares takes tangent; a square minus a constant takes secant. In every case you substitute for x, for dx and for the root, and the identity turns the root into a single trigonometric function.

Around that core sit the practical habits: look for a spare x and an ordinary substitution first, complete the square when the quadratic has an x term, convert the limits of a definite integral instead of converting back, and keep the absolute value on the secant's left branch until you know the sign.

Section 3.4 turns to rational functions and partial fractions. Some of the pieces it produces are one over a sum of squares, whose antiderivative is an inverse tangent, the same triangle you drew today.