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Calculus 2 : Solving Integrals by Substitution

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Example Questions

Example Question #601 : Finding Integrals

Evaluate the following integral:

$\int (\frac{\sqrt{\sec^2(x)-1}}{\cos(x)})dx$

Possible Answers:

$\sec(x)\tan(x)+C$

$\frac{-\sec^3(x)}{3}+C$

$-\sec(x)+C$

$(\sec(x))^{\frac{1}{2}}+\tan(x)+C$

$\sec(x)+C$

Correct answer:

$\sec(x)+C$

Explanation:

To evaluate the integral, we first must rewrite it as the following:

$\int \frac{\sqrt{\tan^2(x)}}{\cos(x)}dx=\int \frac{\tan(x)}{\cos(x)}dx=\int \frac{\sin(x)}{\cos^2(x)}dx$

Now, perform the following subsitution:

$u=\cos(x), du=-\sin(x)$

Next, rewrite the integral and integrate:

$-\int \frac{du}{u^2}=\frac{1}{u}+C$

The integral was performed using the following rule:

$\int x^n dx=\frac{x^{n+1}}{n+1}+C$

Finally, replace with the term:

$\sec(x)+C$

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Example Question #602 : Finding Integrals

Evaluate the following integral:

$\int \frac{dx}{5x^2+1}$

Possible Answers:

$\frac{1}{\sqrt{5}}{\arctan(\sqrt{5}x)}+C$

$\arctan(\frac{x}{\sqrt{5}})+C$

$\frac{1}{5}\arcsin(\sqrt{5}x)+C$

$\frac{1}{5}\arctan(\sqrt{5}x)+C$

Correct answer:

$\frac{1}{\sqrt{5}}{\arctan(\sqrt{5}x)}+C$

Explanation:

To evaluate the integral, we must first make the following substitution:

$u=\sqrt{5}x, du=\sqrt{5}dx$

Now, rewrite the integral and integrate:

$\frac{1}{\sqrt{5}}\int \frac{du}{u^2+1}=\frac{1}{\sqrt{5}}\arctan(u)+C$

The integral was performed using the following rule:

$\int \frac{dx}{x^2+a^2}=\frac{1}{a}\arctan(\frac{x}{a})+C$

Finally, replace u with the x term:

$\frac{1}{\sqrt{5}}{\arctan(\sqrt{5}x)}+C$

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Example Question #23 : Solving Integrals By Substitution

Evaluate the following integral:

$\int \frac{dx}{9x^2+4}$

Possible Answers:

$\arctan(\frac{3x}{2})+C$

$\frac{1}{3}\arctan(\frac{3x}{2})+C$

$\frac{1}{6}\arctan(3x)+C$

$\frac{1}{6}\arctan(\frac{3x}{2})+C$

Correct answer:

$\frac{1}{6}\arctan(\frac{3x}{2})+C$

Explanation:

To integrate, we must first make the following substitution:

u=3x, du=3dx

Now, rewrite the integral in terms of u, and integrate:

$\frac{1}{3}\int \frac{du}{u^2+4}=\frac{1}{6}\arctan(\frac{u}{2})+C$

The integration was performed using the following rule:

$\int \frac{dx}{x^2+a^2}=\frac{1}{a}\arctan(\frac{x}{a})+C$

Finally, replace u with our original term:

$\frac{1}{6}\arctan(\frac{3x}{2})+C$

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Example Question #603 : Finding Integrals

Evaluate the following integral:

$\int \cos^3(x)\sin(x)dx$

Possible Answers:

$\frac{\cos^4(x)\sin^2(x)}{8}+C$

$-\frac{\cos^4(x)}{4}+C$

$\frac{\sin^4(x)}{4}+C$

$\frac{\cos^4(x)}{4}+C$

Correct answer:

$-\frac{\cos^4(x)}{4}+C$

Explanation:

To evaluate the integral, we first must make the following substitution:

$u=\cos(x), du=-\sin(x)dx$

Now, rewrite the integral, and integrate:

$-\int u^3du=-\frac{u^4}{4}+C$

We used the following rule for integration:

$\int x^n dx=\frac{x^{n+1}}{n+1}+C$

Finally, replace with our original term:

$-\frac{\cos^4(x)}{4}+C$

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Example Question #21 : Solving Integrals By Substitution

Evaluate the following integral:

$\int \frac{dx}{\sqrt{x+3}}$

Possible Answers:

$\arctan(x+3)+C$

$\frac{\ln\sqrt{x+3}}{2}+C$

$\ln\sqrt{x+3}+C$

$2\ln\sqrt{x+3}+C$

Correct answer:

$2\ln\sqrt{x+3}+C$

Explanation:

To evaluate the integral, first we must make the following substitution:

$u=\sqrt{x+3}, du=\frac{1}{2}(x+3)^{-\frac{1}{2}}$

The derivative was found using the following rule:

$\frac{\mathrm{d} }{\mathrm{d} x} x^n=nx^{n-1}$

Now, rewrite the integral and integrate:

$2\int \frac{du}{u}=2\ln\left | u\right |+C$

The integral was performed using the following rule:

$\int \frac{dx}{x}=\ln\left | x\right |+C$

Finally, replace with the containing term:

$2\ln\sqrt{x+3}+C$

Note that we removed the absolute value sign because the output of a square root is always positive.

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Example Question #21 : Solving Integrals By Substitution

Evaluate the following integral:

$\int (x^{\frac{3}{2}}+x\sec(3x^2))dx$

Possible Answers:

$x^{\frac{5}{2}}+\ln\left | \sec(x)+\tan(x)\right |+C$

$\frac{2}{5}x^{\frac{5}{2}}+\frac{1}{6}\ln\left | \sec(3x^2)+\tan(3x^2)\right |+C$

$x^{\frac{5}{2}}+6\ln\left | \sec(x)+\tan(x)\right |+C$

$\frac{2}{5}x^{\frac{5}{2}}+\frac{\ln\left | \sec(x)+\tan(x)\right |}{6}+C$

Correct answer:

$\frac{2}{5}x^{\frac{5}{2}}+\frac{1}{6}\ln\left | \sec(3x^2)+\tan(3x^2)\right |+C$

Explanation:

To evaluate the integral, we must split it into two integrals:

$\int x^{\frac{5}{2}}dx+\int x\sec(3x^2)dx$

The first integral is equal to

$\frac{2}{5}x^{\frac{5}{2}}+C$

and was found using the following rule:

$\int x^n dx=\frac{x^{n+1}}{n+1}+C$

The second integral is solved by performing the following substitution:

u=3x^2, du=6xdx

Now, rewrite the integral and integrate:

$\frac{1}{6}\int \sec(u)du=\frac{1}{6}\ln\left | \sec(u)+\tan(u)\right |+C$

The integration was performed using the following rule:

$\int \sec(x)dx=\ln\left | \sec(x)+\tan(x)\right |+C$

Finally, replace with our original term and add the two results of the integrations together:

$\frac{2}{5}x^{\frac{5}{2}}+\frac{1}{6}\ln\left | \sec(3x^2)+\tan(3x^2)\right |+C$

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Example Question #611 : Finding Integrals

Evaluate the following indefinite integral using the substitution method.

$\int 3(x\sec(x^3))^2 dx$

Possible Answers:

$3\sec^2(x^3)+C$

$\tan(\frac{x^4}{4})+C$

$\tan(x^3) + C$

$x^3\tan(x^3)+C$

$\sec^3(x^3)+C$

Correct answer:

$\tan(x^3) + C$

Explanation:

The integral can be expanded by distributing the exponent.

$\int3(x\sec(x^3))^2dx=\int3x^2\sec^2(x^3)dx$

We will make the following substitution:

u=x^3 .

Differentiating both sides yields

du=3x^2dx .

We can then substitute the left hand side of each equation into our integral and evaluate it now.

$\int3x^2\sec^2(x^3)dx=\int\sec^2(u)du=\tan(u)+C.$

Finally, we substitute the original variable back into the expression:

$\tan(u)+C=\tan(x^3)+C$ .

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Example Question #21 : Solving Integrals By Substitution

Solve:

$\int_{0}^{\pi} sin(x)cos(x)dx$

Possible Answers:

$\frac{1}{8}$

$\frac{1}{2}$

$\frac{1}{4}$

Correct answer:

Explanation:

Use substitution:

u = sin(x) , du = cos(x)dx

Plug the and into the regular equation, but no need to worry about the bounds yet:

$\int udu$ $= \frac{1}{2}u^{2}$

Plug back into the integrated equation from above and evaluate from to $\pi$ .

$= \frac{1}{2} (sin(x))^2$

$\frac{1}{2} (sin(\pi)^2 - sin(0)^2) = 0$

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Example Question #961 : Integrals

Solve:

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

Possible Answers:

$\frac{1}{4} (ln(x))^4+C$

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

$\frac{1}{3} (ln(x))^3+C$

$\frac{1}{3} (x)^3+C$

None of the chocies.

Correct answer:

$\frac{1}{3} (ln(x))^3+C$

Explanation:

Use substitution integration:

u = ln(x)

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

$\int u^2 du=\frac{1}{3}u^3+C=\frac{1}{3}(ln(x))^3+C$

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Example Question #22 : Solving Integrals By Substitution

What is the integral of $xe^{x^2}$ ?

Possible Answers:

$\frac{e^{x^2}}{2x}$

$e^{x^2}$

$\frac{e^{x^2}}{x}$

$\frac{1}{2}e^{x^2}$

Correct answer:

$\frac{1}{2}e^{x^2}$

Explanation:

Use substitution:

u = x^2

du = 2xdx

$\frac{du}{2} = xdx$

$\int \frac{e^{u}}{2}du$ $= \frac{e^{u}}{2}$ $= \frac{e^{x^2}}{2}$

Substitute back in.

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Calculus 2 : Solving Integrals by Substitution

Study concepts, example questions & explanations for Calculus 2

All Calculus 2 Resources

Example Questions

Example Question #601 : Finding Integrals

Example Question #602 : Finding Integrals

Example Question #23 : Solving Integrals By Substitution

Example Question #603 : Finding Integrals

Example Question #21 : Solving Integrals By Substitution

Example Question #21 : Solving Integrals By Substitution

Example Question #611 : Finding Integrals

Example Question #21 : Solving Integrals By Substitution

Example Question #961 : Integrals

Example Question #22 : Solving Integrals By Substitution

All Calculus 2 Resources

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