The three inverse trigonometric integration formulas and their general forms with a constant, reaching them by substitution, completing the square for a quadratic denominator, the domains on which each is valid, and telling them apart from the logarithmic forms they resemble.
Subject: Calculus I · 65 slides · symbolic lesson
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Title
Calculus I · Chapter 5 — Integration
Integrals Resulting in Inverse Trigonometric Functions
Objectives
Five outcomes. The formulas are three; the difficulty is recognising which integrands lead to them.
OpenStax Calculus Volume 1, §5.7 Integrals Resulting in Inverse Trigonometric Functions §5.7, pp. 526-532 — the section these objectives are drawn from
Warm-up
Section 3.7 differentiated the inverse trigonometric functions, and the results were surprisingly algebraic — no trigonometry survived in them.
Discussion prompt
The derivative of the inverse tangent is one over one plus x squared. What does that give when read backwards?
Hint: Reverse the statement.
Answer:
\[ \frac{d}{dx}\arctan x = \frac{1}{1+x^{2}} \;\Longrightarrow\; \int\frac{dx}{1+x^{2}} = \arctan x + C \]
So a purely algebraic integrand — a rational function with no trigonometry anywhere — has a trigonometric antiderivative. That is unexpected and it is the section's recurring theme: these formulas are the ones that connect rational-looking integrands to trigonometric answers.
Section 5.6 met exactly this when comparing two quotients with the same denominator. One gave a logarithm and the other could not be done with that section's formulas. This section supplies the missing one.
Concept
Three integrands with no trigonometry in them integrate to inverse trigonometric functions. Recognising those shapes, and reaching them by substitution or by completing the square, is the section's whole content.
the inverse trigonometric forms — Integrands involving the square root of a difference of squares, a sum of squares in a denominator, or the square root of a difference the other way round, each producing an inverse trigonometric function.
\[ \int\frac{dx}{\sqrt{a^{2}-x^{2}}} = \arcsin\frac{x}{a}+C, \qquad \int\frac{dx}{a^{2}+x^{2}} = \frac1a\arctan\frac{x}{a}+C \]
The connection runs both ways: Chapter 6's arc length and the trigonometric substitution technique of later courses both exploit it, and it is why circles and trigonometry keep appearing in problems that started algebraically.
Figure (svg): The three inverse trigonometric integration formulas, with their general forms
OpenStax Calculus Volume 1, §5.7 Integrals Resulting in Inverse Trigonometric Functions §5.7, pp. 575-583
Section
Section 1
Concept
Each inverse trigonometric derivative gives an integration formula, and a substitution generalises each to carry an arbitrary positive constant.
the general forms — Each basic formula extends to one with a constant by substituting the variable divided by that constant, which rescales the integrand and introduces a factor for two of the three.
\[ \int\frac{dx}{a^{2}+x^{2}} = \frac{1}{a}\arctan\frac{x}{a}+C \]
The arctangent form picks up a factor of one over the constant and the arcsine form does not, which is worth noticing rather than memorising: the difference comes from how the constant scales out of each integrand.
Figure (svg): The three inverse trigonometric integration formulas, with their general forms
OpenStax Calculus Volume 1, §5.7 Integrals Resulting in Inverse Trigonometric Functions §5.7, pp. 575-582 — integrals resulting in inverse trigonometric functions
Picture it
The basic forms and their generalisations.
Figure (svg): The three inverse trigonometric integration formulas, with their general forms
The lower three follow from the upper three by one substitution each. Deriving them when needed is quicker and safer than recalling which carries the extra factor.
Worked example
Example 5.58. One substitution generalises the formula.
\[ \text{Derive } \int\frac{dx}{a^{2}+x^{2}} \text{ from the basic form.} \]
Factor the constant out of the denominator
Why: To match the basic form.
\[ a ^{2}(1 + (\frac{x}{a}) ^{2}) \]
Substitute the ratio
Why: The natural choice.
\[ u = \frac{x}{a}, \,dx = a \,du \]
Rewrite
Why: Constants collected.
\[ (a / a ^{2}) \int \,du / (1 + u ^{2}) \]
Integrate
Why: The basic form.
\[ (\frac{1}{a}) \arctan u \]
Substitute back
Why: Undo.
\[ (\frac{1}{a}) \arctan(\frac{x}{a}) + C \]
Figure (svg): The three inverse trigonometric integration formulas, with their general forms
\[ \frac1a\arctan\frac{x}{a}+C \]
Verify: check the arcsine form does not pick up the factor
Why: Doing the same for the arcsine gives a from the differential and a from the square root, which cancel — leaving no external factor. So the arctangent carries one over a and the arcsine carries nothing, and the difference is traceable to the square root rather than being arbitrary. Deriving each takes three lines and removes the need to remember which is which.
OpenStax Calculus Volume 1, §5.7 Integrals Resulting in Inverse Trigonometric Functions §5.7, pp. 578-579
Fill the middle
An arctangent integral with a constant.
Fill in the blanks
\int\frac2___} = \frac______}\arctan\frac______+C
Why: The constant is 2, and the arctangent form carries one over it as an external factor. The arcsine form, by contrast, carries none.
Worked example
Checkpoint 5.58. The general form applied.
\[ \text{Evaluate } \int_{0}^{3/2}\frac{dx}{\sqrt{9-x^{2}}}. \]
Identify the constant
Why: The square root of 9.
\[ a = 3 \]
Apply the general form
Why: No external factor.
\[ \arcsin(\frac{x}{3}) \]
Evaluate at the upper limit
Why: The ratio is one half.
\[ \arcsin(\frac{1}{2}) = \frac{\pi}{6} \]
Evaluate at the lower limit
Why: The ratio is zero.
\[ 0 \]
Subtract
Why: Upper minus lower.
\[ \frac{\pi}{6} \]
Figure (svg): The solution to Worked example a definite arcsine integral shown as a ladder of expressions, one row per legal move
\[ \int_{0}^{3/2}\frac{dx}{\sqrt{9-x^{2}}} = \frac{\pi}{6} \]
Verify: check the interval respects the formula's domain
Why: The formula is valid for x strictly between negative 3 and 3, and the interval from 0 to 1.5 sits comfortably inside — so the integral is legitimate. Had the upper limit been 3 itself the integrand would be unbounded there and the integral would need separate treatment. Note that a purely algebraic integrand produced an exact answer involving pi, which is the section's characteristic surprise.
OpenStax Calculus Volume 1, §5.7 Integrals Resulting in Inverse Trigonometric Functions §5.7, pp. 579-581
Trap
\[ \int\frac{dx}{\sqrt{9-x^{2}}} = \frac13\arcsin\frac{x}{3}+C \]
Attach a factor of one over a
Why: The student generalises from the arctangent formula.
\[ \frac{d}{dx}\left[\tfrac13\arcsin\tfrac{x}{3}\right] = \frac{1}{3\sqrt{9-x^{2}}} \ne \text{integrand} \]
The arcsine form carries no external factor, because the constant cancels between the differential and the square root.
\[ \int\frac{dx}{\sqrt{9-x^{2}}} = \arcsin\frac{x}{3}+C \]
Derive the general form rather than recalling which carries a factor
Why: Three lines settle it, and differentiating confirms it.
The habit worth building is differentiating any answer from this section, since the factor is the only thing that commonly goes wrong and one line exposes it.
Matching
The three basic forms.
Match the pairs
Why: The third and fourth show the asymmetry: the arcsine's constant appears only inside, while the arctangent's appears both inside and as an external factor. Deriving either takes three lines.
Sorting
Look at the shape of the denominator.
Sort into buckets
Sort each integrand.
The presence or absence of a square root is the first thing to look at, and then whether the square is being added or subtracted. Those two questions settle which of the three applies.
Prediction
Commit before reasoning.
Predict first
Why does the general arctangent form carry a factor of one over the constant while the arcsine form does not?
Correct: Because the arcsine's square root supplies a cancelling factor.
\[ \text{arcsin: } \frac{a\,du}{a\sqrt{1-u^{2}}}; \qquad \text{arctan: } \frac{a\,du}{a^{2}(1+u^{2})} \]
Why: Substituting the ratio contributes a factor of the constant from the differential in both cases. For the arctangent the denominator contributes the constant squared, leaving one over the constant; for the arcsine the square root contributes just the constant, which cancels the differential's exactly. So the asymmetry is traceable to the square root and is not a fact to memorise separately.
Section
Section 2
Concept
Few integrands appear in standard form. A substitution — often of a square, an exponential, or a scaled variable — brings them there, and the disguise can be considerable.
recognising a disguised form — An integrand becomes a standard form once a suitable inner function is named. The clue is a squared expression under a root or added to a constant, together with its derivative as a factor.
\[ \int\frac{x\,dx}{\sqrt{1-x^{4}}}, \; u=x^{2} \;\Longrightarrow\; \frac12\arcsin(x^{2})+C \]
The most surprising cases have no algebra in them at all. An exponential integrand can produce an inverse tangent, which nothing about the table's appearance would suggest.
Figure (svg): Reaching a standard form by substitution
OpenStax Calculus Volume 1, §5.7 Integrals Resulting in Inverse Trigonometric Functions §5.7, pp. 578-586 — substitution into inverse trigonometric forms
Picture it
Each needs one substitution.
Figure (svg): Reaching a standard form by substitution
The last is the striking one: an exponential over one plus an exponential squared gives an inverse tangent. Recognising the sum-of-squares shape under the disguise is the whole skill.
Worked example
Example 5.60. A square inside a square root.
\[ \text{Evaluate } \int\frac{x\,dx}{\sqrt{1-x^{4}}}. \]
Recognise the shape
Why: The fourth power is a square squared.
\[ 1 - (x ^{2}) ^{2} \]
Choose the substitution
Why: The inner square.
\[ u = x ^{2} \]
Compute the differential
Why: And adjust.
\[ x \,dx = \,du / 2 \]
Rewrite
Why: The standard arcsine form.
\[ (\frac{1}{2}) \int \,du / \sqrt{1 - u ^{2}} \]
Integrate and substitute back
Why: The basic formula.
\[ (\frac{1}{2}) \arcsin(x ^{2}) + C \]
Figure (svg): Reaching a standard form by substitution
\[ \frac12\arcsin(x^{2})+C \]
Verify: differentiate the answer back
Why: The chain rule gives one half times one over the square root of one minus x to the fourth, times 2x — which is x over that root, the original integrand. Note that the factor of x was essential: without it the integral would be one over the square root of one minus x to the fourth, which is an elliptic integral and not elementary. Once again a single factor of x decides everything.
OpenStax Calculus Volume 1, §5.7 Integrals Resulting in Inverse Trigonometric Functions §5.7, pp. 581-583
Fill the middle
A doubled exponent rewritten.
Fill in the blanks
1+e^x = 1+\left(e^___}\right)^___
Why: A doubled exponent is a square, which turns the denominator into a sum of squares and reveals the arctangent form. Spotting this is the only insight the problem requires.
Worked example
Checkpoint 5.60. A thorough disguise.
\[ \text{Evaluate } \int\frac{e^{x}dx}{1+e^{2x}}. \]
Rewrite the denominator
Why: The exponent doubled is a square.
\[ 1 + (e ^{x}) ^{2} \]
Choose the substitution
Why: The exponential itself.
\[ u = e ^{x} \]
Compute the differential
Why: It matches the numerator.
\[ \,du = e ^{x} \,dx \]
Rewrite
Why: The standard arctangent form.
\[ \int \,du / (1 + u ^{2}) \]
Integrate and substitute back
Why: The basic formula.
\[ \arctan(e ^{x}) + C \]
Figure (svg): The solution to Worked example an exponential giving an arctangent shown as a ladder of expressions, one row per legal move
\[ \arctan(e^{x})+C \]
Verify: differentiate back and note how far the disguise went
Why: The chain rule gives one over one plus the exponential squared, times the exponential — matching. What is striking is that an integrand built entirely from exponentials has a trigonometric antiderivative, with no trigonometry visible anywhere in the problem. Recognising that the doubled exponent makes a square is the only step that requires noticing anything, and everything after it is mechanical.
OpenStax Calculus Volume 1, §5.7 Integrals Resulting in Inverse Trigonometric Functions §5.7, pp. 583-585
Error analysis
A student sees an exponential in a denominator.
Annotate
On: \( \int\frac{e^{x}dx}{1+e^{2x}} = \ln\left|1+e^{2x}\right|+C \)
The check is the one from Section 5.6: differentiate the denominator and compare. Here it fails, which is the signal to look for a different structure — and the doubled exponent is what reveals it.
Sorting
Find the squared expression.
Sort into buckets
Sort each integrand by its substitution.
The fourth is the mirror of the second: a trigonometric integrand giving a trigonometric answer, but an INVERSE one. The pattern to look for is the same in every case — a squared expression with its derivative present.
Two truths and a lie
All three are about reaching the forms.
Eliminate the wrong options
Two of these are true. Knock those out and keep the false one.
Survives elimination: B
Why: The survivor is the false one. Every basic formula in this section has a purely algebraic integrand and a trigonometric antiderivative — the inverse tangent comes from a rational function. That mismatch between an integrand's appearance and its answer is the section's characteristic feature.
Prediction
Commit before reasoning.
Predict first
What shape signals that an integrand may lead to an inverse trigonometric function?
Correct: A squared expression combined with a constant, with its derivative present.
\[ 1+e^{2x} = 1+(e^{x})^{2}, \quad 1-x^{4}=1-(x^{2})^{2} \]
Why: All three formulas have that shape: a constant plus a square, a constant minus a square under a root, or a square minus a constant. The square may be well disguised — a fourth power, a doubled exponent, or a squared trigonometric function — but naming it reveals the form. A square root alone signals nothing, since plenty of roots integrate by the power rule, and the derivative must be present as always.
Section
Section 3
Concept
A quadratic denominator with no real roots can always be written as a square plus a positive constant, which is exactly the arctangent's shape. The shift substitution costs nothing.
completing the square — Rewriting a quadratic as a perfect square plus a constant, which converts an arbitrary quadratic denominator into one of the standard inverse trigonometric shapes.
\[ x^{2}+4x+8 = (x+2)^{2}+2^{2} \]
The shift substitution has derivative one, so no adjustment factor appears — which makes this the cheapest substitution in the subject and completing the square correspondingly worthwhile.
Figure (svg): Completing the square to reveal an inverse trigonometric form
OpenStax Calculus Volume 1, §5.7 Integrals Resulting in Inverse Trigonometric Functions §5.7, pp. 583-590 — completing the square
Picture it
A quadratic denominator handled.
Figure (svg): Completing the square to reveal an inverse trigonometric form
The shift contributes nothing to the differential, so once the square is completed the integral is the general arctangent form read off directly.
Worked example
Example 5.62. Complete the square, then read off.
\[ \text{Evaluate } \int\frac{dx}{x^{2}+4x+8}. \]
Complete the square
Why: Half the linear coefficient, squared.
\[ (x + 2) ^{2} + 4 \]
Identify the constant
Why: The square root of 4.
\[ a = 2 \]
Substitute the shift
Why: Its derivative is one.
\[ u = x + 2, \,du = \,dx \]
Apply the general form
Why: With the external factor.
\[ (\frac{1}{2}) \arctan(\frac{u}{2}) \]
Substitute back
Why: Undo the shift.
\[ (\frac{1}{2}) \arctan(\frac{x + 2}{2}) + C \]
Figure (svg): Completing the square to reveal an inverse trigonometric form
\[ \frac12\arctan\frac{x+2}{2}+C \]
Verify: differentiate the answer back
Why: Differentiating gives one half times one over one plus the square of x plus 2 over 2, times one half — and simplifying the compound fraction returns one over x squared plus 4x plus 8. The shift contributed no factor because its derivative is one, which is what makes completing the square so cheap: the only cost is the algebra, and there is no differential to adjust.
OpenStax Calculus Volume 1, §5.7 Integrals Resulting in Inverse Trigonometric Functions §5.7, pp. 586-587
Fill the middle
A quadratic with a positive leading coefficient.
Fill in the blanks
x^4+4x+8 = (x+2)^___+___
Why: Half the linear coefficient is 2, and its square is 4 — so the constant left over is 8 minus 4. That gives a sum of squares with the constant equal to 2.
Worked example
Checkpoint 5.62. The arcsine version.
\[ \text{Evaluate } \int\frac{dx}{\sqrt{-x^{2}+4x}}. \]
Factor out the negative
Why: To complete the square.
\[ -(x ^{2} - 4 x) \]
Complete the square inside
Why: Half of 4, squared.
\[ -((x - 2) ^{2} - 4) \]
Rewrite
Why: Distribute the negative.
\[ 4 - (x - 2) ^{2} \]
Identify the form
Why: The arcsine's shape with a = 2.
Integrate
Why: The general form.
\[ \arcsin(\frac{x - 2}{2}) + C \]
Figure (svg): The solution to Worked example a quadratic under a root shown as a ladder of expressions, one row per legal move
\[ \arcsin\frac{x-2}{2}+C \]
Verify: check the domain the answer implies
Why: The arcsine requires its argument between negative one and one, so x must lie between 0 and 4 — which is exactly where the original radicand is positive, since it factors as x times 4 minus x. The two conditions agree, as they must, and that agreement is a useful check that the completion was done correctly. Note the negative sign had to be factored out before completing the square, which is the step most often mishandled.
OpenStax Calculus Volume 1, §5.7 Integrals Resulting in Inverse Trigonometric Functions §5.7, pp. 587-589
Trap
\[ -x^{2}+4x = (-x+2)^{2}-4 \quad \text{(wrong)} \]
Complete the square without factoring the negative
Why: The student works directly on the negative leading term.
\[ (-x+2)^{2}-4 = x^{2}-4x \ne -x^{2}+4x \]
Expanding shows the sign of the squared term is wrong, so the whole expression is the negative of what was wanted.
\[ -x^{2}+4x = -(x^{2}-4x) = 4-(x-2)^{2} \]
Factor the negative out, complete the square inside, then distribute
Why: The completion formula assumes a positive leading coefficient.
Expanding the result confirms it, which takes one line. Since the sign decides whether the answer is an arcsine or something else entirely, the check is worth making.
Ranking
A quadratic denominator.
Put in order
Why: Step a is the one skipped, and skipping it produces the negative of the intended expression — which changes the form from an arcsine to something with no real answer at all. Expanding the completed square is a one-line check.
Sorting
Look at the resulting shape.
Sort into buckets
Sort each quadratic.
The last is worth checking for first: a factorable quadratic has a different treatment entirely, and completing the square on it produces a square MINUS a constant, which is the arcsecant's shape rather than the arctangent's.
Prediction
Commit before reasoning.
Predict first
Why does substituting x plus a constant require no adjustment factor?
Correct: Because the derivative of a shift is one.
\[ u = x+2 \;\Longrightarrow\; du = dx \quad \text{(nothing to adjust)} \]
Why: Differentiating x plus a constant gives 1, so du equals dx exactly and nothing needs adjusting. That makes a shift the cheapest substitution available and is why completing the square is so effective: all the work is algebraic, and the substitution itself is free. Contrast a scaling substitution, whose derivative is the scale factor and which does contribute one.
Section
Section 4
Concept
Integrands sharing a denominator can give an inverse tangent, a logarithm, or a mixture, depending entirely on the numerator. Checking it is the only reliable way to choose.
distinguishing the forms — Compare the numerator with the denominator's derivative: matching gives a logarithm, a bare constant gives an inverse trigonometric function, and a numerator of equal or higher degree needs division first.
\[ \frac{2x}{x^{2}+1} \to \ln; \qquad \frac{1}{x^{2}+1} \to \arctan \]
The third case is the one most often missed: when the numerator's degree is at least the denominator's, dividing first is mandatory, and the answer then has both a polynomial and a transcendental part.
Figure (svg): Three similar-looking quotients with completely different antiderivatives
OpenStax Calculus Volume 1, §5.7 Integrals Resulting in Inverse Trigonometric Functions §5.7, pp. 580-590 — distinguishing integration forms
Picture it
One denominator throughout.
Figure (svg): Three similar-looking quotients with completely different antiderivatives
The three answers are a transcendental function, a different transcendental function, and a mixture. Nothing about the shared denominator predicts which, and only the numerator does.
Worked example
Example 5.64. Divide before anything else.
\[ \text{Evaluate } \int\frac{x^{2}}{x^{2}+1}\,dx. \]
Compare the degrees
Why: Both are two.
Divide
Why: Add and subtract one on top.
\[ 1 - \frac{1}{x ^{2} + 1} \]
Integrate the first term
Why: A constant.
Integrate the second
Why: The arctangent form.
\[ -\arctan x \]
Combine
Why: With the constant.
\[ x - \arctan x + C \]
Figure (svg): Three similar-looking quotients with completely different antiderivatives
\[ x-\arctan x+C \]
Verify: differentiate the answer back
Why: Differentiating gives 1 minus one over one plus x squared, and combining over a common denominator gives x squared over x squared plus one — the original integrand. The division was mandatory: attempting an inverse trigonometric form directly on an improper quotient fails, since no formula in the table has a numerator of that degree. Checking degrees before choosing a form takes a second.
OpenStax Calculus Volume 1, §5.7 Integrals Resulting in Inverse Trigonometric Functions §5.7, pp. 588-589
Sorting
Look at the numerator.
Sort into buckets
Sort each integrand with denominator x^2 + 1.
The last case is the general one and contains the others: any linear numerator splits into a derivative part and a constant part, giving a logarithm plus an inverse tangent. That decomposition handles every quotient with an irreducible quadratic below.
Worked example
Checkpoint 5.64. Two forms in one integral.
\[ \text{Evaluate } \int\frac{x+3}{x^{2}+1}\,dx. \]
Split the numerator
Why: Two separate fractions.
\[ \frac{x}{x ^{2} + 1} + \frac{3}{x ^{2} + 1} \]
Recognise the first
Why: Matches the derivative up to a factor.
Integrate it
Why: With the factor of one half.
\[ (\frac{1}{2}) \ln(x ^{2} + 1) \]
Recognise the second
Why: A constant on top.
Integrate it and combine
Why: With its coefficient.
\[ +3 \arctan x + C \]
Figure (svg): The solution to Worked example splitting a mixed numerator shown as a ladder of expressions, one row per legal move
\[ \frac12\ln(x^{2}+1)+3\arctan x+C \]
Verify: differentiate back and note the general technique
Why: Differentiating gives x over x squared plus one, plus 3 over the same denominator, which recombines to the original integrand. The technique generalises: any numerator can be split into a part proportional to the denominator's derivative and a constant remainder, giving a logarithm and an inverse tangent respectively. That decomposition is the standard first move for any quotient with an irreducible quadratic denominator.
OpenStax Calculus Volume 1, §5.7 Integrals Resulting in Inverse Trigonometric Functions §5.7, pp. 589-590
Error analysis
A student integrates a quotient of equal degrees.
Annotate
On: \( \int\frac{x^{2}}{x^{2}+1}dx = \arctan x + C \)
Comparing the degrees before choosing a form is the check. When the numerator's degree is at least the denominator's, division is mandatory and the answer will have a polynomial part.
Fill the middle
A numerator of the same degree as the denominator.
Fill in the blanks
\frac1}___+1} = 1 - \frac___}___+1}
Why: Writing the numerator as the denominator minus one splits the fraction into a constant and a standard arctangent form. Division is mandatory whenever the numerator's degree is at least the denominator's.
Two truths and a lie
All three are about telling the forms apart.
Eliminate the wrong options
Two of these are true. Knock those out and keep the false one.
Survives elimination: B
Why: The survivor is the false one. Three integrands sharing the denominator x squared plus one give an inverse tangent, a logarithm, and a mixture — the denominator is identical in all three and predicts nothing. The numerator decides, which is why comparing it with the denominator's derivative is the first move.
Prediction
Commit before reasoning.
Predict first
For any quotient with an irreducible quadratic denominator, what is the standard first step?
Correct: Divide if improper, then split the numerator.
\[ \frac{x+3}{x^{2}+1} = \frac12\cdot\frac{2x}{x^{2}+1}+\frac{3}{x^{2}+1} \]
Why: That decomposition handles every case: the derivative part integrates to a logarithm and the constant part to an inverse tangent, with a polynomial part appearing first if division was needed. Completing the square comes afterwards if the quadratic is not already in standard shape. Assuming a form without looking at the numerator is exactly the error this idea exists to prevent.
Section
Section 5
Concept
The arcsine form requires the radicand positive, the arcsecant form requires the variable outside an interval, and only the arctangent form is valid everywhere. An integral must respect those limits.
domain restrictions — The interval on which an integrand is defined and bounded, and on which its antiderivative formula therefore applies. An integral crossing outside it is not covered by the formula.
\[ \int\frac{dx}{\sqrt{a^{2}-x^{2}}}: \quad -a < x < a \]
The arctangent's integrand is defined and bounded for every input, which is why it appears far more often than the other two and why its integral over an unbounded interval converges.
Figure (svg): Where each formula is valid: the domains of the three inverse functions
OpenStax Calculus Volume 1, §5.7 Integrals Resulting in Inverse Trigonometric Functions §5.7, pp. 576-590 — domains of the formulas
Picture it
Where each formula applies.
Figure (svg): Where each formula is valid: the domains of the three inverse functions
The arcsine integrand becomes unbounded at the ends of its interval, so an integral reaching those endpoints needs the improper-integral treatment rather than a direct evaluation.
Worked example
Example 5.63. The endpoint where the integrand blows up.
\[ \text{What is different about } \int_{0}^{2}\frac{dx}{\sqrt{4-x^{2}}}? \]
Check the integrand at the upper limit
Why: The radicand vanishes.
\[ \text{unbounded at } x = 2 \]
Note the integral is improper
Why: Not a standard definite integral.
\[ \text{Section } 5.2' s\text{ condition fails} \]
Apply the formula anyway
Why: Formally.
\[ \arcsin(\frac{x}{2})\text{ from } 0\text{ to } 2 \]
Evaluate
Why: The arcsine of one.
\[ \frac{\pi}{2} \]
Note why the answer is still valid
Why: The area converges despite the spike.
Figure (svg): Where each formula is valid: the domains of the three inverse functions
\[ \int_{0}^{2}\frac{dx}{\sqrt{4-x^{2}}} = \frac{\pi}{2} \]
Verify: see why an unbounded integrand can still enclose finite area
Why: The integrand grows without bound as x approaches 2, yet the region under it has finite area — the spike is infinitely tall but narrows fast enough. Approaching the limit from just below gives 1.5099 at x equal to 1.99 and 1.5608 at 1.999, closing on pi over 2. Not every unbounded integrand behaves this way: the reciprocal near zero does not, and its integral diverges. Which happens depends on how fast the growth is.
OpenStax Calculus Volume 1, §5.7 Integrals Resulting in Inverse Trigonometric Functions §5.7, pp. 585-586
Sorting
Check the interval against the domain.
Sort into buckets
Sort each integral of 1 over the square root of 4 minus x squared.
The middle category is the interesting one: an unbounded integrand can still enclose a finite area if it narrows fast enough. Whether it does is a genuine question, and Section 5.6's reciprocal near zero is a case where it does not.
Worked example
Checkpoint 5.63. Bounded everywhere.
\[ \text{Evaluate } \int_{-\infty}^{\infty}\frac{dx}{1+x^{2}} \text{ informally.} \]
Note the integrand is bounded
Why: Between 0 and 1 everywhere.
Antidifferentiate
Why: The basic form.
Consider the behaviour at large positive x
Why: The arctangent's asymptote.
\[ \text{approaches } \frac{\pi}{2} \]
Consider large negative x
Why: The other asymptote.
\[ \text{approaches } -\frac{\pi}{2} \]
Subtract
Why: The total.
Figure (svg): Why the arctangent appears so often: a bounded antiderivative
\[ \int_{-\infty}^{\infty}\frac{dx}{1+x^{2}} = \pi \]
Verify: connect the finite area to the bounded antiderivative
Why: The area under this curve across the entire real line is exactly pi, a finite number despite the interval being unbounded — because the integrand decays like one over x squared, fast enough for the tails to contribute little. That finiteness is the same fact as the arctangent having horizontal asymptotes: a bounded antiderivative and a convergent total area are two statements of one thing. This is the shape of every probability distribution's normalisation, and indeed one over pi times this integrand is the Cauchy distribution.
OpenStax Calculus Volume 1, §5.7 Integrals Resulting in Inverse Trigonometric Functions §5.7, pp. 586-588
Trap
\[ \int_{0}^{4}\frac{dx}{\sqrt{4-x^{2}}} = \arcsin\frac{x}{2}\bigg|_{0}^{4} = \arcsin 2 - 0 \]
Substitute a limit beyond the domain
Why: The student applies the formula mechanically.
The arcsine of 2 is undefined, and beyond x equal to 2 the radicand is negative so the integrand is not even real.
\[ \text{the integrand is undefined for } x>2 \]
Check the interval against the integrand's domain before evaluating
Why: A formula says nothing where its integrand does not exist.
Here the error at least announces itself, since the arcsine of 2 is visibly meaningless. Section 5.6's reciprocal across the origin was more dangerous, because there the formula produced a plausible-looking number.
Fill the middle
The arcsine form's requirement.
Fill in the blanks
\int\fraca___-x^___}}: \quad |x| < ___
Why: The radicand must be positive, which requires the variable's magnitude to stay below the constant. Beyond that the integrand is not real, so no formula applies.
Two truths and a lie
All three are about domains.
Eliminate the wrong options
Two of these are true. Knock those out and keep the false one.
Survives elimination: B
Why: The survivor is the false one. The arcsine form requires the variable's magnitude below the constant and the arcsecant form requires it above — each fails outside its interval, where the radicand goes negative. Only the arctangent has no such restriction, which is part of why it appears so much more often.
Prediction
Commit before reasoning.
Predict first
Why does the arctangent form appear far more often than the other two?
Correct: Because its integrand is defined and bounded everywhere.
\[ a^{2}+x^{2} > 0 \text{ always}; \quad a^{2}-x^{2} \text{ changes sign at } \pm a \]
Why: A sum of a positive constant and a square never vanishes and never goes negative, so the integrand has no singularities and no domain restriction — it can be integrated over any interval, including unbounded ones. The other two have radicands that change sign, restricting them to intervals. The absence of a square root is a symptom of this rather than the reason.
Comparison
Fill the blanks. The numerator decides everything.
Comparison matrix
| Numerator | What to do | Answer |
|---|---|---|
| 1 | the arctangent form directly | arctan x |
| 2x | the logarithmic form | ln(x^2+1) |
| x^2 | divide first: it is improper | x - arctan x |
| x + 3 | split into both forms | (1/2) ln(x^2+1) + 3 arctan x |
The last row is the general case and contains the others. Splitting a numerator into a multiple of the denominator's derivative plus a constant handles every quotient with an irreducible quadratic below.
Pattern
Given a quotient or a radical that may lead to an inverse trigonometric function.
Step one is the check most often skipped, and no formula in the table has a numerator of degree equal to the denominator's — so skipping it produces an answer that differentiates to something else entirely.
Stewart, Calculus: Early Transcendentals 8e, §7.3 Trigonometric Substitution §7.3, pp. 486-492
Check
The general forms.
Check your understanding
What is the integral of 1 over the square root of 9 minus x squared?
Answer: A
Why: The arcsine form carries no external factor: the constant cancels between the differential and the root.
Check
Telling the forms apart.
Check your understanding
What is the integral of x squared over x squared plus one?
Answer: A
Why: The quotient is improper, so dividing gives 1 minus one over x squared plus one.
Check
Domains.
Check your understanding
Which of the three forms is valid for every real input?
Answer: A
Why: A sum of a positive constant and a square never vanishes or goes negative.
Real world
A rotating radar dish tracks an aircraft flying past at constant speed along a straight line, at a perpendicular distance of 3 kilometres. The dish's bearing angle changes at a rate the servo motor must supply.
Discussion prompt
Explain why an inverse tangent appears, what integrating the angular rate gives, and why the total swept angle is finite.
Hint: The bearing is determined by a ratio of distances.
Answer:
With the aircraft at horizontal position x from the closest point, the bearing satisfies the tangent of the angle equal to x over 3 — so the angle is the inverse tangent of that ratio. Section 3.7's derivative gives the angular rate, and it is a purely algebraic expression despite describing an angle.
\[ \theta = \arctan\frac{x}{3}, \qquad \frac{d\theta}{dt} = \frac{3}{9+x^{2}}\cdot\frac{dx}{dt} \]
Integrating that rate over an interval gives the angle swept — the Net Change Theorem again — and the integrand is exactly this section's arctangent form with the constant equal to 3.
The total swept angle over the whole flight is finite, and equal to pi: the dish turns through half a revolution as the aircraft passes from far on one side to far on the other, however long the flight. That is the same convergence as the arctangent's horizontal asymptotes, and it is what lets a servo be specified with a bounded total travel rather than an open-ended one.
Note that the angular rate peaks at the closest approach and falls off like one over the square of the distance, which is why tracking is hardest directly overhead — a fact the integral's shape predicts before any numbers are put in.
Commit first
Answer, then rate your confidence honestly.
Predict first
Three integrands share the denominator x squared plus one. What decides which formula applies?
Correct: The numerator.
\[ \frac{2x}{x^{2}+1}\to\ln, \quad \frac{1}{x^{2}+1}\to\arctan, \quad \frac{x^{2}}{x^{2}+1}\to x-\arctan x \]
Why: With 2x on top the answer is a logarithm, with 1 an inverse tangent, and with x squared a mixture of a polynomial and an inverse tangent — three unrelated answers from one denominator. The check is to compare the numerator with the denominator's derivative and to compare degrees, both of which take a line. Assuming a form because the denominator looks familiar is the section's characteristic error.
Explain it
They wrote that the integral of x squared over x squared plus one is the inverse tangent.
Discussion prompt
In four sentences or fewer, show them the check.
Hint: Have them differentiate their answer.
Answer:
Ask them to differentiate the inverse tangent: it gives one over x squared plus one, not x squared over it — so the numerator is wrong. No formula in the table has a numerator of the same degree as its denominator, which is the signal that division is needed first.
Dividing gives 1 minus one over x squared plus one, and integrating that gives x minus the inverse tangent. Comparing the degrees before choosing a form is a one-second habit that catches this every time.
Exit ticket
Name the weakest spot before you close the deck.
Predict first
Which of these would you least want handed to you cold?
Correct: Whichever you picked is tonight's ten minutes, and each has a one-line fix.
Why: For the formulas, derive the general forms rather than recalling which carries a factor. For substitution, look for a squared expression with its derivative present. For completing the square, factor out any negative first and expand to check. For telling them apart, compare the numerator with the denominator's derivative and compare degrees. Do five of your chosen kind rather than twenty mixed ones.
Connect it up
Paper. Fifteen minutes.
Draw it
At the top, write the three basic formulas with their general forms beside them, marking which carries an external factor and deriving that one in three lines. Below, write the three quotients sharing the denominator x squared plus one, with their unrelated answers, and one sentence on what decides which. In the middle of the page, complete the square on a quadratic denominator in five lines through to the final answer, and beside it do the same for one with a negative leading coefficient, circling the step where the negative is factored out. In the lower half, draw the three domains as intervals on a number line with the arcsine integrand sketched above its interval, showing where it blows up. At the bottom, sketch the arctangent's integrand and antiderivative side by side, mark the two horizontal asymptotes, and write one line connecting the finite total area to the bounded antiderivative.
If your two sketches at the bottom do not visibly connect — the shaded area on one to the asymptote height on the other — redraw them, because that connection is what makes the total area of pi memorable rather than a fact to recall.
Recap
Five things, and the fourth is the check this section exists to install.
| If you see | Then |
|---|---|
| A constant minus a square, under a root | Arcsine, with no external factor |
| A constant plus a square below | Arctangent, with one over the constant |
| A doubled exponent or a fourth power | Substitute to expose the square |
| A quadratic denominator | Complete the square first |
| A negative leading coefficient | Factor it out before completing |
| A numerator matching the derivative | A logarithm, not an inverse tangent |
| A numerator of equal degree | Divide before anything else |
That completes Chapter 5's formula list and the chapter itself. Chapter 6 stops adding techniques and starts spending them: areas between curves, volumes, arc length, work and centres of mass, all built on the definite integral this chapter defined and learned to compute.
OpenStax Calculus Volume 1, §5.7 Integrals Resulting in Inverse Trigonometric Functions §5.7, pp. 526-532 — everything on these slides traces back here
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