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Calculus 2 : Finding Integrals

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Example Question #96 : Solving Integrals By Substitution

Calculate the following integral:

$y=\int \sqrt{4-4x^{2}}dx$

Possible Answers:

$y=sin^{-1}x+x\sqrt{1-x^2}+C$

$y=sin^{-1}x-sin(2sin^{-1}x)+C$

$y=2sin^{-1}x-sin(2sin^{-1}x)+C$

$y=2sin^{-1}x-\frac{1}{2}sin(2sin^{-1}x)+C$

Correct answer:

$y=sin^{-1}x+x\sqrt{1-x^2}+C$

Explanation:

$y=\int \sqrt{4-4x^{2}}dx$

Factor out $\sqrt{4}$ from the integrand, and simplify:

$y=\int \sqrt{4}\sqrt{1-x^{2}}dx=\int 2\sqrt{1-x^{2}}dx=2\int \sqrt{1-x^{2}}dx$

Make the following substitution: $x=sin\theta$ $dx=cos\theta d\theta$

Plug the substitution into the integrand:

$y=2\int \sqrt{1-sin^{2}\theta }cos\theta d\theta$ .

Use the Pythagorean identity to make the following substitution, and simplify:

$\\y=2\int \sqrt{1-sin^{2}\theta }cos\theta d\theta\\=2\int \sqrt{cos^{2}\theta }cos\theta d\theta\\=2\int cos\theta cos\theta d\theta\\= 2\int cos^{2}\theta d\theta$

Apply the following identity to the integrand:

$cos^{2}\theta=\frac{1+cos(2 \theta )}{2}$ :

$y=2\int cos^{2}\theta=2\int \frac{1+cos(2) \theta }{2}d\theta$ .

Separate the integral into two separate integrals:

$y= 2\int \frac{1}{2}d\theta+2\int\frac{cos(2\theta )}{2}d\theta$ .

Solve the first integral:

$y=\theta+2\int\frac{cos(2\theta )}{2}d\theta$ .

Make the following substitution for the second integral:

$u=2\theta$ $du=2d\theta$ $\frac{1}{2}du=d\theta$ .

Apply the substitution, and solve the integral:

$2\int \frac{cos(2\theta )}{2}=2\int \frac{cosu}{2}\frac{1}{2}du=\frac{1}{2}sinu=\frac{1}{2}sin(2\theta )+C$ .

Combine answers for both integrals:

$y=2\int cos^{2}\theta=\theta+\frac{1}{2}sin(2\theta )+C$

Solve for $\theta$ :

$x=sin\theta$

$sin^{-1}x=\theta$

Plug values for $\theta$ back into solution to integral:

Recall that,

$sin(2\theta)=2sin(\theta)cos(\theta)$

and from above,

$\\x=sin(\theta) \\cos(\theta)=\sqrt{1-sin^2(\theta)}=\sqrt{1-x^2}$

Therefore,

$\\\theta+\frac{1}{2}sin(2\theta )+C \\=sin^{-1}+4sin(x)cos(x) \\=sin^{-1}(x)+x\sqrt{1-x^2}+C$ .

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Example Question #97 : Solving Integrals By Substitution

Evaluate the integral:

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

Possible Answers:

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

2sin(x)+C

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

2sin(x^2)+C

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

Correct answer:

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

Explanation:

You must use u substitution to solve this problem.

In this case

$u=\frac{1}{2}x^2$

and

du=xdx .

So the integral simplifies to

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

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Example Question #98 : Solving Integrals By Substitution

Calculate the following integral:

$\int_{1}^{9}x\sqrt[3]{x-1}dx$

Possible Answers:

384/7

468/7

572

125/4

Correct answer:

468/7

Explanation:

Use $t=\sqrt[3]{x-1}$ as a substitute, then:

$t^3=x-1\:\Rightarrow\:x=t^3+1\:\Rightarrow\:dx=3t^2dt$ ;

Now, rewrite the boundaries of integration in terms of t:

$\newline t_1=\sqrt[3]{x_1-1}=\sqrt[3]{1-1}=0;\newline t_2=\sqrt[3]{x_2-1}=\sqrt[3]{9-1}=2;$

Rewrite the integral in terms of t:

$\\\int_{1}^{9}x\sqrt[3]{x-1}dx\\\\=\int_{0}^{2}(t^3+1)t\:3t^2\:dt\\\\=3\int_{0}^{2}t^6\:dt +3\int_{0}^{2}t^3\:dt\\\\=\frac{3*128}{7}+\frac{3*16}{4}\\\\=\frac{468}{7}$

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Example Question #99 : Solving Integrals By Substitution

Evaluate the following integral:

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

Possible Answers:

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

$-e^{\frac{1}{x}}+c$

$\frac{1}{e^x}+c$

$e^{\frac{1}{x}}+c$

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

Correct answer:

$-e^{\frac{1}{x}}+c$

Explanation:

We must use substitution to solve this integrand.

Let

$u=\frac{1}{x}$

and

$du=-\frac{1}{x^2}dx$

don't forget to divide buy -1 to isolate the right side

$-du=\frac{1}{x^2}dx$

putting all the values back in and pulling the negative sign out of the integral we get:

$\int \frac{e^{\frac{1}{x}}}{x^2}dx=-\int e^udu=-e^u+c=-e^{\frac{1}{x}}+c$

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Calculus 2 : Finding Integrals

Study concepts, example questions & explanations for Calculus 2

All Calculus 2 Resources

Example Questions

Example Question #96 : Solving Integrals By Substitution

Example Question #97 : Solving Integrals By Substitution

Example Question #98 : Solving Integrals By Substitution

Example Question #99 : Solving Integrals By Substitution

All Calculus 2 Resources

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