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Calculus 2 : Integrals

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Example Questions

Example Question #11 : Solving Integrals By Substitution

Evaluate the integral:

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

Possible Answers:

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

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

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

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

Correct answer:

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

Explanation:

To evaluate the integral, we must recognize that what we were given looks very similar to the following integral:

$\int \frac{dx}{\sqrt{a^2-x^2}}=\arcsin(\frac{x}{a})+C$

To make our integral look like the one above, we must perform the following substiution:

u=3x, du=3dx

Now, rewrite our integral:

$\frac{1}{3}\int \frac{du}{\sqrt{4-u^2}}$

It looks like the one above, so we can integrate now:

$\frac{1}{3}\int \frac{du}{\sqrt{4-u^2}} = \frac{1}{3}\arcsin(\frac{u}{2})+C$

Finally, replace u with our original term:

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

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Example Question #12 : Solving Integrals By Substitution

Evaluate the following integral:

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

Possible Answers:

$\frac{{x^5}}{5}+\ln(x)+\arctan(2x)+C$

$\frac{x^5}{5}+\ln(x)+\arcsin(2x)+C$

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

$\frac{x^5}{5}-\ln(x)+\arcsin(2x)+C$

$\frac{x^5}{5}+\ln(x)+\frac{\arctan(2x)}{2}+C$

Correct answer:

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

Explanation:

To integrate, we must break the integral into three integrals:

$\int x^4dx +\int \frac{dx}{x}+\int \frac{dx}{4x^2+`1}$

The first integral is equal to

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

and was found using the following rule:

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

The second integral is equal to

$\ln\left | x\right |+C$

and was found using the following rule:

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

The final integral is found by performing the following substitution:

u=2x, du=2dx

Now, rewrite and integrate:

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

The integral was found using the following rule:

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

Finally, rewrite the integral in terms of by replacing with the original term, and add all three integrals together to get a final answer of

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

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Example Question #13 : Solving Integrals By Substitution

Evaluate the following integral:

$\int \cos^3(x)\sin^5(x)dx$

Possible Answers:

$\frac{\sin^8(x)}{8}-\frac{\cos^6(x)}{6}+C$

$-\frac{\sin^8(x)}{8}+\frac{\sin^6(x)}{6}+C$

$\sin^8(x)-\sin^6(x)+C$

$\frac{\sin^8(x)}{8}-\frac{\sin^6(x)}{6}+C$

Correct answer:

$-\frac{\sin^8(x)}{8}+\frac{\sin^6(x)}{6}+C$

Explanation:

To integrate, we must make the following substitution:

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

The derivative was found using the following rule:

$\frac{\mathrm{d} }{\mathrm{d} x} \sin(x)=\cos(x)$

Now, rewrite the integral:

$\int (1-u^2)(u^5)du$

Notice that we changed

$\cos^2(x)=1-\sin^2(x)=1-u^2$

Next, distribute and integrate:

$\int (u^5-u^7)du=\frac{u^6}{6}-\frac{u^8}{8}+C$

The integral was found using the following rule:

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

Finally, replace with our original term:

$-\frac{\sin^8(x)}{8}+\frac{\sin^6(x)}{6}+C$

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Example Question #14 : Solving Integrals By Substitution

Evaluate the following integral:

$\int \cos(x)\sin^6(x)dx$

Possible Answers:

$\frac{\cos^7(x)}{7}+C$

$\frac{\sin^7(x)}{7}+C$

$\sin(x)\cos(x)+C$

Correct answer:

$\frac{\sin^7(x)}{7}+C$

Explanation:

To solve the integral, we must make the following substitution:

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

Now, rewrite the integral and integrate:

$\int u^6 du=\frac{u^7}{7} + C$

The integral was found using the following rule:

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

Finally, replace with our original term:

$\frac{\sin^7(x)}{7}+C$

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Example Question #21 : Solving Integrals By Substitution

Evaluate the following integral:

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

Possible Answers:

$-\sec(x)+C$

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

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

$\sec(x)+C$

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

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 #22 : Solving Integrals By Substitution

Evaluate the following integral:

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

Possible Answers:

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

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

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

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

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 #21 : Solving Integrals By Substitution

Evaluate the following integral:

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

Possible Answers:

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

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

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

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

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 #23 : Solving Integrals By Substitution

Evaluate the following integral:

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

Possible Answers:

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

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

$\arctan(x+3)+C$

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

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 #26 : Solving Integrals By Substitution

Evaluate the following integral:

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

Possible Answers:

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

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

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

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

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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Calculus 2 : Integrals

Study concepts, example questions & explanations for Calculus 2

All Calculus 2 Resources

Example Questions

Example Question #11 : Solving Integrals By Substitution

Example Question #12 : Solving Integrals By Substitution

Example Question #13 : Solving Integrals By Substitution

Example Question #14 : Solving Integrals By Substitution

Example Question #21 : Solving Integrals By Substitution

Example Question #22 : Solving Integrals By Substitution

Example Question #23 : Solving Integrals By Substitution

Example Question #21 : Solving Integrals By Substitution

Example Question #23 : Solving Integrals By Substitution

Example Question #26 : Solving Integrals By Substitution

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