Module 21: Trigonometric integrals and trigonometric substitution

8 exercises, each with a complete solution. 1 figure.

On this page

A. Prerequisites and observable goals
B. The central idea: use identities to expose a derivative pattern
C. Recognizable patterns and branch choices
D. Why the substitutions remove the radical
E. Choose the least complicated substitution
F. Worked examples
G. Independent practice
H. Complete solutions: Module 21
I. Retention and readiness

A. Prerequisites and observable goals#

Before

Trigonometric identities, substitution, inverse trigonometry, and absolute values.

After

Choose a parity-based trig integration strategy and use a trig substitution while preserving its interval and square-root sign.

B. The central idea: use identities to expose a derivative pattern#

Trigonometric integration reorganizes products until one factor is the derivative of a useful substitution. Trigonometric substitution instead changes a radical expression into a trigonometric identity. The first starts with trigonometric functions; the second often starts with an algebraic square root. Both are substitutions justified by identities, not unrelated collections of tricks.

The substitution $x=a\sin\theta$ as a triangle. In a right triangle with hypotenuse

a

and angle

\theta

, the opposite side is

x=a\sin\theta

and the adjacent side is

\sqrt{a^2-x^2}=a\cos\theta

. The radical becomes a single trigonometric factor, and the same triangle converts the answer back to

x

C. Recognizable patterns and branch choices#

For $\sin^m x\cos^n x$ , an odd sine power can reserve one $\sin x\dd x$ and turn the remaining sine squares into $1-\cos^2x$ . An odd cosine power works symmetrically. When both powers are even, use

\sin^2x=\frac{1-\cos2x}{2},\qquad \cos^2x=\frac{1+\cos2x}{2}.

For tangent/secant products, an even secant power can reserve $\sec^2x\dd x$ for $u=\tan x$ ; an odd tangent power together with a secant factor may reserve $\sec x\tan x\dd x$ for $u=\sec x$ . Cases outside these patterns can need identities, parts, or another approach.

For $a>0$ , the standard substitutions are

\sqrt{a^2-x^2}:x=a\sin\theta;\quad \sqrt{a^2+x^2}:x=a\tan\theta;\quad \sqrt{x^2-a^2}:x=a\sec\theta.

Choose the $\theta$ interval explicitly. For example, $x=a\sin\theta$ , $-\pi/2\le\theta\le\pi/2$ , ensures $\sqrt{a^2\cos^2\theta}=a\cos\theta$ because cosine is nonnegative there. Without that branch condition the square root is $a|\cos\theta|$ . For $x>a$ with secant substitution, $0<\theta<\pi/2$ is convenient; a negative- $x$ branch needs separate treatment. [O2Trig]

D. Why the substitutions remove the radical#

The Pythagorean identities $1-\sin^2\theta=\cos^2\theta$ , $1+\tan^2\theta=\sec^2\theta$ , and $\sec^2\theta-1=\tan^2\theta$ turn a difference or sum of squares into a single square. The differential also changes: $x=a\tan\theta$ gives $\dd x=a\sec^2\theta\dd\theta$ . Omitting this Jacobian factor changes the integral.

After integration, a right-triangle relation or inverse trig identity returns the result to $x$ . A triangle is a memory aid on the chosen branch, not permission to ignore signs for negative inputs. Different-looking primitives can agree up to a constant on their common interval; differentiation is the decisive check.

E. Choose the least complicated substitution#

Try an ordinary substitution before a trig substitution. For $\int x\sqrt{x^2+9}\dd x$ , the nearby $x\dd x$ makes $u=x^2+9$ simpler. Reserve trig substitution for radicals whose remaining factors do not already match a straightforward derivative. With definite bounds, transform both endpoints and maintain their order. With indefinite integrals, convert every trig term back before the final answer.

F. Worked examples#

Odd sine power. For $\int\sin^3x\cos^2x\dd x$ , reserve one sine and write $\sin^2x=1-\cos^2x$ . Let $u=\cos x$ , $\dd u=-\sin x\dd x$ :

-\int(1-u^2)u^2\dd u=-u^3/3+u^5/5+C.

The answer is $-\cos^3x/3+\cos^5x/5+C$ ; the negative sign comes from the differential.

A circle radical. For $\int\sqrt{9-x^2}\dd x$ on $(-3,3)$ , use $x=3\sin\theta$ on the principal branch. The integrand times differential becomes $9\cos^2\theta\dd\theta$ . Half-angle integration yields $(9/2)(\theta+\sin\theta\cos\theta)+C$ , hence

\frac92\arcsin(x/3)+\frac{x\sqrt{9-x^2}}2+C.

This primitive will produce familiar circular areas at appropriate bounds.

G. Independent practice#

F foundation C standard course work T transfer S optional stretch L runnable lab or project. Attempt the complete set before opening its solutions.

M21.01FSolution

Evaluate $\int\sin x\cos^4x\dd x$ .

M21.02CSolution

Evaluate $\int_0^{\pi/2}\sin^2x\dd x$ .

M21.03CSolution

Evaluate $\int\tan^3x\sec^2x\dd x$ on an interval avoiding secant singularities.

M21.04TSolution

Evaluate $\int\sec^3x\tan x\dd x$ .

M21.05CSolution

Evaluate $\int\dd x/\sqrt{16-x^2}$ on $(-4,4)$ , stating a trig-substitution branch.

M21.06TSolution

Evaluate $\int\dd x/(x^2+4)^{3/2}$ using trigonometric substitution.

M21.07TSolution

For $x>3$ , evaluate $\int\dd x/[x\sqrt{x^2-9}]$ using $x=3\sec\theta$ .

M21.08SSolution

Evaluate $\int_0^2 x^2/\sqrt{4-x^2}\dd x$ as an improper limit. Explain why the endpoint singularity does not force divergence.

Optional small hints

For M21.04, reserve $\sec x\tan x\dd x$ . For M21.06, choose $x=2\tan\theta$ with $-\pi/2<\theta<\pi/2$ . For M21.08, replace the upper endpoint by $b<2$ , then take $b\to2^-$ .

H. Complete solutions: Module 21#

Compare the first step where your work differs. A corrected answer is not yet an independent success.

M21.01Show solution

The differential supplies the sign

Let $u=\cos x$ , so $\dd u=-\sin x\dd x$ . The integral is $-\int u^4\dd u=-u^5/5+C$ , or $\boxed{-\cos^5x/5+C}$ . Its derivative is $\sin x\cos^4x$ .

Back to M21.01

M21.02Show solution

An even power calls for a half-angle identity

Using $\sin^2x=(1-\cos2x)/2$ gives $\int_0^{\pi/2}\sin^2x\dd x=[x/2-\sin2x/4]_0^{\pi/2}=\boxed{\pi/4}$ . It is half the interval length, consistent with the complementary equal integrals of sine squared and cosine squared.

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M21.03Show solution

Reserve the derivative of tangent

Let $u=\tan x$ and $\dd u=\sec^2x\dd x$ . The integral is $u^4/4+C=\boxed{\tan^4x/4+C}$ on each interval avoiding $\cos x=0$ . The primitive does not define an antiderivative across those singularities.

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M21.04Show solution

Reserve the derivative of secant

Write the integrand as $\sec^2x(\sec x\tan x)$ . With $u=\sec x$ , it becomes $\int u^2\dd u$ , giving $\boxed{\sec^3x/3+C}$ . This is a different derivative pattern from M21.03 despite the similar symbols.

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M21.05Show solution

Choose a branch before dropping an absolute value

Set $x=4\sin\theta$ , $-\pi/2<\theta<\pi/2$ . Then $\dd x=4\cos\theta\dd\theta$ and $\sqrt{16-x^2}=4\cos\theta>0$ . Their ratio is $\dd\theta$ , so the primitive is $\boxed{\arcsin(x/4)+C}$ . Differentiating gives $(1/4)/\sqrt{1-x^2/16}=1/\sqrt{16-x^2}$ on the stated interval.

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M21.06Show solution

The differential and denominator powers both matter

With $x=2\tan\theta$ on the principal tangent interval, $\dd x=2\sec^2\theta\dd\theta$ and $(x^2+4)^{3/2}=8\sec^3\theta$ , since secant is positive. Thus the integral is $\tfrac14\int\cos\theta\dd\theta=\sin\theta/4+C$ . The relation $\sin\theta=x/\sqrt{x^2+4}$ gives $\boxed{x/(4\sqrt{x^2+4})+C}$ . Differentiation simplifies to $(x^2+4)^{-3/2}$ for all real $x$ .

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M21.07Show solution

Restrict secant to the specified positive branch

For $x>3$ , choose $0<\theta<\pi/2$ . Then $\dd x=3\sec\theta\tan\theta\dd\theta$ and $\sqrt{x^2-9}=3\tan\theta$ . The integral becomes $\tfrac13\int\dd\theta$ , hence $\boxed{\tfrac13\arccos(3/x)+C}$ on $(3,\infty)$ . Checking by chain rule gives $(1/3)(3/x^2)/\sqrt{1-9/x^2}=1/(x\sqrt{x^2-9})$ , using $x>0$ .

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M21.08Show solution

An unbounded height may still have finite accumulated area

For $b<2$ , substitute $x=2\sin\theta$ , taking the transformed upper bound $\beta=\arcsin(b/2)$ . The quotient times differential becomes $4\sin^2\theta\dd\theta$ . Thus the truncated integral is $[2\theta-\sin2\theta]_0^\beta$ . As $b\to2^-$ , $\beta\to\pi/2$ , so the limit is $\boxed\pi$ . The integrand is unbounded at the endpoint, but the defining improper limit is finite; singularity alone does not decide convergence.

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I. Retention and readiness#

Retrieve M04.08, M01.06, and M17.05. Explain why $\sqrt{a^2\cos^2\theta}$ includes an absolute value until the branch is chosen. Advance when you can decide between an ordinary substitution and a trigonometric one without relying only on the presence of a radical.