Calculus 1 Quiz: Selecting Integration Techniques

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Selecting Integration TechniquesQuestion 1 of 20

To evaluate $\int \ln(x^2+1) dx$ , the first step is to use integration by parts. What integral remains to be solved after this first step?

\int \frac{2x}{x^2+1} dx

\int \frac{x^2}{x^2+1} dx

\int \frac{2x^2}{x^2+1} dx

\int x \ln(x^2+1) dx

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Calculus 1 Quiz

Calculus 1 Quiz: Selecting Integration Techniques

Practice Selecting Integration Techniques in Calculus 1 with focused quiz questions that help you check what you know, review explanations, and build confidence with test-style prompts.

What this quiz covers

This quiz focuses on Selecting Integration Techniques, giving you a quick way to practice the rules, question types, and explanations that matter most for Calculus 1.

How to use this quiz

Try each quiz question before looking at the correct answer. Use the explanations to review missed ideas, then come back to similar questions until the pattern feels familiar.

All questions

Question 1

To evaluate $\int \ln(x^2+1) dx$ , the first step is to use integration by parts. What integral remains to be solved after this first step?

$\int \frac{2x}{x^2+1} dx$
$\int \frac{x^2}{x^2+1} dx$
$\int \frac{2x^2}{x^2+1} dx$ (correct answer)
$\int x \ln(x^2+1) dx$

Explanation: For

\int \ln(x^2+1) dx

, we use integration by parts with

u = \ln(x^2+1)

and

dv = dx

. This gives

du = \frac{2x}{x^2+1} dx

and

v = x

. The integration by parts formula,

\int u dv = uv - \int v du

, yields:

x \ln(x^2+1) - \int x \cdot \frac{2x}{x^2+1} dx

. The remaining integral is

\int \frac{2x^2}{x^2+1} dx

Question 2

To evaluate $\int \sec^3(x) dx$ , a common method involves integration by parts. What is the most effective choice for $u$ and $dv$ ?

$u = \sec^3(x)$ , $dv = dx$
$u = \sec(x)$ , $dv = \sec^2(x) dx$ (correct answer)
$u = \sec^2(x)$ , $dv = \sec(x) dx$
$u = 1$ , $dv = \sec^3(x) dx$

Explanation: The standard approach for

\int \sec^n(x) dx

(for odd

n

) is integration by parts. We need to choose

dv

to be something we can easily integrate. By splitting

\sec^3(x)

\sec(x) \cdot \sec^2(x)

, we can choose

u = \sec(x)

and

dv = \sec^2(x) dx

. This is effective because

du = \sec(x)\tan(x) dx

is manageable and

v = \int \sec^2(x) dx = \tan(x)

is a basic integral. This choice leads to a solvable integral using a trigonometric identity.

Question 3

Which of the following integrals requires polynomial long division as a necessary preliminary step?

$\int \frac{x^2-x+6}{x^3+3x} dx$
$\int \frac{x^3+2x^2-x+1}{x^2+1} dx$ (correct answer)
$\int \frac{x+4}{x^2+5x+6} dx$
$\int \frac{1}{x^4-1} dx$

Explanation: Polynomial long division is required when the degree of the numerator is greater than or equal to the degree of the denominator in a rational function. In choice B, the degree of the numerator is 3 and the degree of the denominator is 2. Therefore, one must divide

x^3+2x^2-x+1

x^2+1

before proceeding with integration techniques like partial fractions on the remainder term. The other options all have numerators with a strictly smaller degree than their denominators.

Question 4

For the integral $\int \frac{dx}{x+\sqrt{x}}$ what is the most productive initial substitution?

A trigonometric substitution with $x=\tan^2\theta$ .
A rationalizing substitution with $u=\sqrt{x}$ . (correct answer)
A simple substitution with $u=x+\sqrt{x}$ .
Integration by parts with $u=\frac{1}{x+\sqrt{x}}$ and $dv=dx$ .

Explanation: The presence of terms with different powers of

x

, including a root, suggests a rationalizing substitution. Let

u = \sqrt{x}

. Then

x = u^2

and

dx = 2u \, du

. Substituting these into the integral gives

\int \frac{2u \, du}{u^2+u} = \int \frac{2u}{u(u+1)} du = \int \frac{2}{u+1} du

. This is a simple logarithmic integral, demonstrating the effectiveness of the substitution.

Question 5

A student is asked to evaluate $\int \frac{1}{1+e^x} dx$ . A helpful first step is to multiply the numerator and denominator by $e^{-x}$ . What technique should be applied immediately after this algebraic step?

Integration by parts, letting $u=e^{-x}$ .
Partial fraction decomposition on the new denominator.
A u-substitution, letting $u$ equal the new denominator. (correct answer)
A trigonometric substitution, treating $e^{-x/2}$ as $\tan(\theta)$ .

Explanation: Multiplying the numerator and denominator by

e^{-x}

transforms the integral into

\int \frac{e^{-x}}{e^{-x}+1} dx

. This form is now perfectly set up for a u-substitution. If we let

u = e^{-x}+1

, then

du = -e^{-x} dx

. The integral becomes

\int \frac{-du}{u} = -\ln|u| + C

, which can then be substituted back in terms of

x

Question 6

The integral $\int \frac{(x^2 + 2)^2}{x^4} \,dx$ appears complex. Which of the following is the most direct method to find its antiderivative?

Use the substitution $u = x^2 + 2$ .
Use integration by parts with $u = (x^2+2)^2$ .
Simplify the integrand by algebraic expansion and division before integrating. (correct answer)
Use a trigonometric substitution with $x = \sqrt{2}\tan(\theta)$ .

Explanation: The correct answer is C. Before attempting complex integration techniques, algebraic simplification should always be considered. By expanding the numerator, we get:

\frac{(x^2 + 2)^2}{x^4} = \frac{x^4 + 4x^2 + 4}{x^4}

. Dividing each term by

x^4

gives

1 + \frac{4}{x^2} + \frac{4}{x^4} = 1 + 4x^{-2} + 4x^{-4}

. This expression can be easily integrated term-by-term using the power rule. A is incorrect because the substitution

u = x^2+2

leads to

du = 2x \,dx

, and it's difficult to account for the

1/x^4

term. B is incorrect because integration by parts would be very cumbersome and complicated. D is incorrect because trigonometric substitution is not necessary and would be far more work than algebraic simplification.

Question 7

Which integral's antiderivative is found most directly by applying a standard inverse trigonometric function formula, such as those for arcsin or arctan?

$\int \frac{x}{1+x^4} \,dx$
$\int \frac{1}{x^2+2x+5} \,dx$
$\int \frac{1}{\sqrt{9-x^2}} \,dx$ (correct answer)
$\int \arctan(x) \,dx$

Explanation: The correct answer is C. The integral

\int \frac{1}{\sqrt{9-x^2}} \,dx

matches the standard form

\int \frac{1}{\sqrt{a^2-u^2}} \,du = \arcsin(\frac{u}{a}) + C

, with

a=3

and

u=x

. No preliminary steps are needed. A is incorrect because it requires a u-substitution (

u=x^2

) before it fits the arctan form. B is incorrect because it requires completing the square in the denominator to get

\int \frac{1}{(x+1)^2+4} \,dx

before it fits the arctan form. D is incorrect because this integral is solved using integration by parts, where

\arctan(x)

is part of the integrand, not the result of a standard formula.

Question 8

Which integral is best evaluated by first splitting it into two separate integrals, which then require two different antidifferentiation techniques?

$\int \frac{x+1}{x^2+1} \,dx$ (correct answer)
$\int (x^2 + \sin(x)) \,dx$
$\int x(x^2+1)^3 \,dx$
$\int (\ln(x))^2 \frac{1}{x} \,dx$

Explanation: When you encounter an integral that looks complex, always ask yourself: "Can I split this into simpler parts that require different techniques?" This strategy is particularly useful when the numerator of a rational function contains terms that suggest different approaches. Looking at option A,

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

, you can split this as

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

. The first integral requires substitution (let

u = x^2+1

), giving you

\frac{1}{2}\ln(x^2+1)

. The second integral is a standard arctangent form, giving you

\arctan(x)

. Two completely different techniques are needed. Option B is incorrect because

\int (x^2 + \sin(x)) \,dx

splits into two integrals that both use basic power and trigonometric rules - the same fundamental antidifferentiation technique category. Option C is wrong because

\int x(x^2+1)^3 \,dx

is best handled as one integral using substitution (

u = x^2+1

), not by splitting it up. Option D is incorrect because

\int (\ln(x))^2 \frac{1}{x} \,dx

should be tackled with a single substitution (

u = \ln(x)

), making it

\int u^2 \,du

. Study tip: When you see a sum in a numerator over a single denominator, try splitting it into separate fractions. If the resulting integrals require noticeably different techniques (like substitution versus standard forms), you've found the right approach.

Question 9

For an integral of the form $\int \frac{P(x)}{Q(x)} \,dx$ , where $P(x)$ and $Q(x)$ are polynomials and the degree of $P(x)$ is greater than or equal to the degree of $Q(x)$ , what is the necessary first step before applying techniques like partial fraction decomposition?

Use integration by parts with $u=P(x)$ and $dv = \frac{1}{Q(x)} \,dx$ .
Factor the denominator $Q(x)$ into linear and irreducible quadratic factors.
Perform polynomial long division to rewrite the integrand. (correct answer)
Apply substitution with $u = Q(x)$ to simplify the rational expression.

Explanation: The correct answer is C. When integrating a rational function where the degree of the numerator is greater than or equal to the degree of the denominator (an improper rational function), the required first step is to perform polynomial long division. This will rewrite the integrand as the sum of a polynomial and a proper rational function (where the numerator's degree is less than the denominator's). The polynomial part can be integrated easily, and techniques like partial fraction decomposition can then be applied to the proper rational function remainder. A is incorrect; integration by parts is not the standard approach for rational functions. B is a necessary step for partial fraction decomposition, but it must come after long division if the fraction is improper. D is incorrect; substitution with

u = Q(x)

would not simplify this type of rational expression effectively.

Question 10

To evaluate $\int \frac{1}{x^2 - 6x + 13} dx$ , what is the most appropriate sequence of techniques?

Use partial fraction decomposition by factoring the denominator, then integrate each term.
Complete the square in the denominator, then apply a u-substitution that leads to an arctangent form. (correct answer)
Apply the substitution $u = x^2 - 6x + 13$ , which transforms the integral into a standard logarithmic form.
Use a trigonometric substitution with $x = \sqrt{13} \sec(\theta)$ to simplify the expression in the denominator.

Explanation: The denominator

x^2 - 6x + 13

is an irreducible quadratic (discriminant

b^2-4ac = 36-52 < 0

), so partial fractions are not applicable. The correct approach is to complete the square:

x^2 - 6x + 9 + 4 = (x-3)^2 + 2^2

. The integral becomes

\int \frac{1}{(x-3)^2 + 2^2} dx

. A u-substitution

u=x-3

then yields an integral of the form

\int \frac{1}{u^2+a^2}du

, which results in an arctangent function.

Question 11

The integral $\int (\ln(x) + 1) \,dx$ can be solved more efficiently than term-by-term integration by recognizing the integrand as the result of a specific derivative rule. Which rule is it?

The chain rule applied to $\ln(x^2)$ .
The product rule applied to $x\ln(x)$ . (correct answer)
The quotient rule applied to $\frac{\ln(x)}{x}$ .
The sum rule applied to $\int \frac{1}{x} \,dx$ and $\int \ln(x) \,dx$ .

Explanation: The correct answer is B. By the product rule, the derivative of

x\ln(x)

\frac{d}{dx}(x\ln(x)) = (1)(\ln(x)) + (x)(\frac{1}{x}) = \ln(x) + 1

. Therefore,

\int (\ln(x) + 1) \,dx = x\ln(x) + C

. Recognizing this pattern is much more efficient than integrating term-by-term. A is incorrect;

\frac{d}{dx}(\ln(x^2)) = \frac{2x}{x^2} = \frac{2}{x}

. C is incorrect;

\frac{d}{dx}(\frac{\ln(x)}{x}) = \frac{\frac{1}{x} \cdot x - \ln(x) \cdot 1}{x^2} = \frac{1-\ln(x)}{x^2}

. D describes the term-by-term approach, which is valid but less efficient as it requires using integration by parts for

\int \ln(x) \,dx

Question 12

Which of the following integrals is most appropriately solved using integration by parts rather than a direct u-substitution?

$\int x \sin(x^2) \,dx$
$\int x \sin(x) \,dx$ (correct answer)
$\int \frac{\sin(x)}{\cos^2(x)} \,dx$
$\int \frac{\sin(x)}{1 + \cos(x)} \,dx$

Explanation: The correct answer is B. Integration by parts is typically used for integrals of products of functions from different 'classes' (e.g., polynomial times trigonometric). In

\int x \sin(x) \,dx

, we have a polynomial (

x

) and a trigonometric function (

\sin(x)

). Choosing

u=x

and

dv = \sin(x) \,dx

leads to a simpler integral. A is incorrect because it is a straightforward u-substitution problem. Let

u = x^2

, then

du = 2x \,dx

. The integral becomes

\frac{1}{2} \int \sin(u) \,du

. C is incorrect because it can be solved with u-substitution. Let

u = \cos(x)

, then

du = -\sin(x) \,dx

. The integral becomes

-\int u^{-2} \,du

. Alternatively, one could rewrite it as

\int \sec(x)\tan(x) \,dx

. D is incorrect because it is a u-substitution problem. Let

u = 1 + \cos(x)

, then

du = -\sin(x) \,dx

. The integral becomes

-\int \frac{1}{u} \,du

Question 13

When using integration by parts to evaluate $\int x^3 \ln(x) \,dx$ , what is the most strategic choice for $u$ and $dv$ ?

$u = x^3$ and $dv = \ln(x) \,dx$
$u = \ln(x)$ and $dv = x^3 \,dx$ (correct answer)
$u = x$ and $dv = x^2 \ln(x) \,dx$
$u = 1$ and $dv = x^3 \ln(x) \,dx$

Explanation: The correct answer is B. The formula for integration by parts is

\int u \,dv = uv - \int v \,du

. The goal is to choose

u

and

dv

such that the new integral,

\int v \,du

, is simpler than the original. By choosing

u = \ln(x)

and

dv = x^3 \,dx

, we get

du = \frac{1}{x} \,dx

and

v = \frac{x^4}{4}

. The new integral is

\int \frac{x^4}{4} \cdot \frac{1}{x} \,dx = \int \frac{x^3}{4} \,dx

, which is a simple power rule integral. A is incorrect because choosing

dv = \ln(x) \,dx

requires finding the integral of

\ln(x)

, which itself requires integration by parts, making the process unnecessarily complex. C is incorrect because finding

v

from

dv = x^2 \ln(x) \,dx

is more difficult than the original problem. D is incorrect because choosing

dv

as the entire integrand just returns the original problem in the

\int v \,du

term, making no progress.

Question 14

To evaluate $\int \frac{2x-1}{x^2-x-6} \,dx$ , a student observes that the denominator factors into $(x-3)(x+2)$ . Which statement describes the most efficient solution strategy?

Use the substitution $u=x^2-x-6$ , since the numerator is the derivative of the denominator. (correct answer)
Use partial fraction decomposition to write the integrand as $\frac{A}{x-3} + \frac{B}{x+2}$ .
Use integration by parts with $u = 2x-1$ and $dv = \frac{1}{x^2-x-6} \,dx$ .
Use a trigonometric substitution because the denominator is a quadratic expression.

Explanation: When you encounter a rational function where the numerator might be related to the derivative of the denominator, always check this relationship first—it often leads to the most efficient solution path. Let's examine the key insight: the denominator is

x^2 - x - 6

, and its derivative is

2x - 1

. Notice that the numerator is exactly

2x - 1

! This means we can use the substitution

u = x^2 - x - 6

, making

du = (2x - 1)dx

. The integral becomes

\int \frac{1}{u} du = \ln|u| + C = \ln|x^2 - x - 6| + C

. This is remarkably clean and direct. Answer A is correct because it recognizes this derivative relationship, leading to an immediate logarithmic antiderivative. Answer B, partial fractions, would work but requires more steps: finding constants A and B, then integrating two separate fractions. While mathematically valid, it's unnecessarily complex when the simpler substitution is available. Answer C suggests integration by parts, but this technique is designed for products of functions, not rational functions—and choosing

dv = \frac{1}{x^2-x-6} dx

would make the problem harder, not easier. Answer D mentions trigonometric substitution, which applies to expressions like

\sqrt{a^2 - x^2}

or similar forms involving square roots, not rational functions with factored denominators. Study tip: Before diving into partial fractions for rational integrals, always check if the numerator is the derivative (or a constant multiple) of the denominator. This simple check can save you significant time and algebraic work.

Question 15

Consider the definite integral $\int_{-\pi/2}^{\pi/2} (x^2 \sin(x) + \cos(x)) \,dx$ . What is the most efficient method to evaluate this integral?

Utilize the property of odd functions over a symmetric interval $[ -a, a ]$ to simplify the integral before finding an antiderivative. (correct answer)
Find the antiderivative of each term, using integration by parts for $x^2 \sin(x)$ , and then apply the Fundamental Theorem of Calculus.
Apply the substitution $u = \sin(x)$ to the first term to simplify the integral.
Use a numerical method like the Trapezoidal Rule, as the antiderivative is too complex to find by hand.

Explanation: When you encounter a definite integral over a symmetric interval like

[-\pi/2, \pi/2]

, always check if the integrand contains odd or even functions. This can dramatically simplify your work. Let's examine each term in

\int_{-\pi/2}^{\pi/2} (x^2 \sin(x) + \cos(x)) \,dx

. The function

x^2 \sin(x)

is the product of

x^2

(even) and

\sin(x)

(odd), making it an odd function overall. Since

f(-x) = (-x)^2 \sin(-x) = x^2(-\sin(x)) = -x^2\sin(x) = -f(x)

, this confirms it's odd. For any odd function integrated over a symmetric interval

[-a,a]

, the result is always zero because the positive and negative areas cancel perfectly. The second term,

\cos(x)

, is an even function. So our integral becomes:

0 + \int_{-\pi/2}^{\pi/2} \cos(x) \,dx = 2\int_{0}^{\pi/2} \cos(x) \,dx = 2[\sin(x)]_0^{\pi/2} = 2(1-0) = 2

. Choice A correctly identifies this efficient approach. Choice B would work but requires the tedious integration by parts of

x^2\sin(x)

, which is unnecessarily complex. Choice C suggests substitution, but

u = \sin(x)

doesn't effectively simplify

x^2\sin(x)

. Choice D incorrectly assumes the antiderivative is too difficult when symmetry properties make it trivial. Strategy tip: Always check for odd/even function properties when you see symmetric integration limits. This single observation can turn a difficult problem into a simple one.

Question 16

For the integral $\int x^3 e^{x^2} \,dx$ , a student considers the substitution $u = x^2$ . How should the student proceed after making this substitution?

The substitution transforms the integral into $\int u e^u \,du$ , which can be solved with integration by parts.
The substitution fails because the integrand contains an odd power of $x$ that cannot be fully expressed in terms of $u$ .
The integral transforms directly into an elementary antiderivative involving $e^u$ without further techniques.
The integral transforms into $\frac{1}{2} \int u e^u \,du$ , which can then be solved using integration by parts. (correct answer)

Explanation: When you encounter an integral like

\int x^3 e^{x^2} \,dx

, substitution is often the right approach, but you need to carefully track how all parts of the integrand transform. Let's work through the substitution

u = x^2

systematically. First, find

du = 2x \,dx

, so

x \,dx = \frac{1}{2} du

. Now you need to express

x^3

in terms of

u

. Since

u = x^2

, you have

x^3 = x \cdot x^2 = x \cdot u

. Substituting everything:

\int x^3 e^{x^2} \,dx = \int (xu) e^u \cdot \frac{1}{2x} \,du = \frac{1}{2} \int u e^u \,du

. This integral requires integration by parts with

v = u

and

dw = e^u du

. Looking at the wrong answers: Choice A omits the crucial factor of

\frac{1}{2}

that comes from the

du

substitution. Choice B incorrectly claims the substitution fails—while

x^3

contains an odd power, we can still express it in terms of

u

and the remaining

x

factor. Choice C suggests the integral becomes elementary, but

\int u e^u \,du

definitely requires integration by parts, not a simple antiderivative formula. The key insight is that successful substitution doesn't just mean the exponential simplifies—you must account for how

dx

transforms and ensure all remaining

x

terms can be expressed using your substitution variable. Always double-check that every part of your integrand converts properly before proceeding.

Question 17

To evaluate $\int e^{2x} \cos(x) \,dx$ , integration by parts is required. Which statement accurately describes the full procedure?

Apply integration by parts once with $u = e^{2x}$ to directly find the antiderivative.
This integral cannot be expressed in terms of elementary functions and requires numerical approximation.
A u-substitution with $u = 2x$ should be performed before using integration by parts, which then solves the integral in one step.
Apply integration by parts twice, which results in the original integral appearing on one side of an equation, allowing it to be solved for algebraically. (correct answer)

Explanation: When you encounter an integral like

\int e^{2x} \cos(x) \,dx

that involves the product of an exponential and trigonometric function, integration by parts is your go-to technique. However, this particular integral requires a special approach that creates a clever algebraic loop. Using integration by parts with

u = e^{2x}

and

dv = \cos(x)dx

, you get

e^{2x}\sin(x) - 2\int e^{2x}\sin(x)dx

. Now you must apply integration by parts again to the new integral

\int e^{2x}\sin(x)dx

. Setting

u = e^{2x}

and

dv = \sin(x)dx

gives you

-e^{2x}\cos(x) + 2\int e^{2x}\cos(x)dx

. Notice that your original integral has reappeared! Substituting back, you get:

\int e^{2x}\cos(x)dx = e^{2x}\sin(x) - 2[-e^{2x}\cos(x) + 2\int e^{2x}\cos(x)dx]

. This simplifies to an equation where your original integral appears on both sides, allowing you to solve algebraically for it. Choice A is wrong because one application of integration by parts doesn't complete the solution—it creates another integral requiring the same technique. Choice B is incorrect since this integral does have an elementary antiderivative. Choice C fails because the substitution

u = 2x

doesn't simplify the integral structure, and integration by parts still requires the two-step process. Study tip: When integrating products of exponentials and trig functions, expect to use integration by parts twice to create an algebraic equation you can solve for the original integral.

Question 18

To evaluate the integral $\int \frac{x^3 + 4x}{x^2 + 2} dx$ , which of the following is the most effective initial step?

Use the substitution $u = x^2 + 2$ , which transforms the integral but requires further manipulation of the remaining $x^2$ term.
Apply integration by parts with $u = x^3 + 4x$ and $dv = \frac{1}{x^2 + 2} dx$ , leading to a more complex integral involving $\arctan(x)$ .
Perform algebraic simplification by factoring an $x$ from the numerator and recognizing that $x^2+2$ is not a factor.
Use polynomial long division or algebraic manipulation to rewrite the integrand as $x + \frac{2x}{x^2 + 2}$ . (correct answer)

Explanation: The degree of the numerator (3) is greater than the degree of the denominator (2), so the first step should be to simplify the rational expression. By long division, or by rewriting

x^3+4x = x(x^2+2) + 2x

, we get

\frac{x^3+4x}{x^2+2} = \frac{x(x^2+2)+2x}{x^2+2} = x + \frac{2x}{x^2+2}

. The integral then becomes

\int (x + \frac{2x}{x^2+2}) dx

, which can be solved easily. The first term is a simple power rule, and the second term is solved with a u-substitution.

Question 19

What is the most effective substitution to begin evaluating the integral $\int \frac{x^3}{\sqrt{x^2+4}} dx$ ?

A u-substitution, $u = x^2+4$ , because its derivative $2x$ is related to a factor in the numerator.
A trigonometric substitution, $x = 2 \tan(\theta)$ , which simplifies the radical term $\sqrt{x^2+4}$ . (correct answer)
Integration by parts, with $u = x^2$ and $dv = \frac{x}{\sqrt{x^2+4}} dx$ , simplifying the polynomial part.
A rationalizing substitution, $u = \sqrt{x^2+4}$ , to eliminate the radical from the expression entirely.

Explanation: The presence of the term

\sqrt{x^2+a^2}

(here

a=2

) strongly suggests a trigonometric substitution. Setting

x = 2 \tan(\theta)

transforms

\sqrt{x^2+4}

into

\sqrt{4\tan^2(\theta)+4} = 2\sec(\theta)

. The integral becomes

\int \frac{8\tan^3(\theta)}{2\sec(\theta)} (2\sec^2(\theta)) d\theta = 8\int \tan^3(\theta)\sec(\theta) d\theta

, which is a standard trigonometric integral. While other methods like (A) or (C) are possible, they are more complex and less direct than trigonometric substitution for this form.

Question 20

For the integral $\int \sin^5(x) \cos^2(x) dx$ , which strategy is the most direct path to a solution?

Convert all terms to $\sin(x)$ using $\cos^2(x) = 1 - \sin^2(x)$ and then integrate powers of sine.
Use half-angle identities to reduce the powers of both the sine and cosine terms before integrating.
Split off a factor of $\sin(x)$ and use $\sin^2(x) = 1 - \cos^2(x)$ to prepare for a $u = \cos(x)$ substitution. (correct answer)
Apply integration by parts with $u = \sin^5(x)$ and $dv = \cos^2(x) dx$ , which requires using a half-angle identity.

Explanation: For integrals of the form

\int \sin^m(x) \cos^n(x) dx

, if at least one of the powers (m or n) is odd, the standard strategy is to split off one factor from the odd power and convert the remaining even power using Pythagorean identities. Here, the power of sine is odd (5). We write

\int \sin^4(x) \cos^2(x) \sin(x) dx = \int (\sin^2(x))^2 \cos^2(x) \sin(x) dx = \int (1-\cos^2(x))^2 \cos^2(x) \sin(x) dx

. Now, the substitution

u = \cos(x)

du = -\sin(x) dx

transforms the integral into

-\int (1-u^2)^2 u^2 du

, which is a polynomial integral.