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Calculus 2 : Integrals

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Example Questions

Example Question #2 : Solving Integrals By Substitution

Solve the following indefinite integral:

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

Possible Answers:

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

$2(1-4x^2)^{\frac{1}{2}}+C$

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

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

Correct answer:

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

Explanation:

To solve the following integral, we must make a substitution to create the following general form:

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

We make the following subsitution:

u=2x, du=2dx

The derivative was found using the following rule:

$\frac{d}{dx}(x^n)=nx^{n-1}$

The integral now looks like this:

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

Notice that it is in the same form as the integral we want.

Now use the form from above to integrate:

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

To finish the problem, replace u with 2x:

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

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Example Question #4 : Solving Integrals By Substitution

Simplify the following indefinite integral.

$\small \int (\ln x)^2 dx$

Possible Answers:

$\small e^x(x^2-2x+2)+C$

$\small e^x((\ln x)^2-2\ln x+2)+C$

$\small \small \small x((\ln x )^2-2\ln x+2)+C$

$\small \small \small \small x((\ln x )^2+2\ln x-2)+C$

Correct answer:

$\small \small \small x((\ln x )^2-2\ln x+2)+C$

Explanation:

We can simplify

$\small \int (\ln x)^2 dx$

by first doing a substitution, with $\small u=\ln x$ , which gives us $\small e^u=x$ , which means that $\small e^udu=dx$ . So the integral becomes

$\small \small \int (\ln x)^2 dx=\int u^2e^u du$

The integral $\small \small \int u^2 e^u du$ can be solved using two integration by parts, which will give us

$\small \small \small \small \int u^2 e^u du=e^u(u^2-2u+2)+C$

So now we just plg in $\small \ln x$ into $\small u$ and $\small x$ into $\small e^u$ to get

$\small \small \small \small \int (\ln x)^2 dx=\int u^2e^u du=e^u(u^2-2u+2)+C=x((\ln x )^2-2\ln x+2)+C$

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Example Question #9 : Solving Integrals By Substitution

Calculate the integral: $\int_{2}^{6}\frac{2}{3xln^3(x)} \:dx$

Possible Answers:

$\frac{1}{3ln^2(6)}-\frac{1}{3ln^2(2)}$

$\frac{1}{3ln^2(2)}-\frac{1}{3ln^2(6)}$

$\frac{2}{3ln^2(6)}-\frac{2}{3ln^2(2)}$

$\frac{2}{3ln^2(2)}-\frac{2}{3ln^2(6)}$

$\frac{1}{3ln^2(2)}+\frac{1}{3ln^2(6)}$

Correct answer:

$\frac{1}{3ln^2(2)}-\frac{1}{3ln^2(6)}$

Explanation:

Pull out the constant out in front of the integral.

$\int_{2}^{6}\frac{2}{3xln^3(x)} \:dx=\frac{2}{3}\int_{2}^{6}\frac{1}{xln^3(x)} \:dx$

Use U-substitution to solve.

u=ln(x)

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

$\frac{2}{3}\int_{2}^{6}\frac{1}{xln^3(x)} \:dx= \frac{2}{3}\int u^{-3}\:du= \frac{2}{3}\left[ \frac{u^{-2}}{-2}\right]=-\frac{1}{3}\left[\frac{1}{ln^2(u)}\right]_{2}^{6}$

$-\frac{1}{3}\left[\frac{1}{ln^2(6)}-\frac{1}{ln^2(2)}\right]=\frac{1}{3ln^2(2)}-\frac{1}{3ln^2(6)}$

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Example Question #3 : Solving Integrals By Substitution

Solve the following indefinite integral:

$\int \frac{dx}{\sqrt{2x+3}}$

Possible Answers:

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

$\ln \left | \sqrt{2x+3} \right |+C$

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

$\tan^{-1}(2x+3)+C$

Correct answer:

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

Explanation:

The integral can be solved with a clever substitution:

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

The derivative was found using the following rules:

$\frac{d}{dx}(x^n)=nx^{n-1}$ , $\frac{d}{dx}f(g(x))=f'(g(x))\cdot g'(x)$

Then, when you rewrite the integral in terms of u, you find that you get:

$\int du = u+C$

The integration was performed using the following rule:

$\int dx=x+C$

Finally, replace the u with our original term.

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

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Example Question #11 : Solving Integrals By Substitution

Solve the following indefinite integral:

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

Possible Answers:

$\tan (x+2)+C$

$\ln ((x+2)^2+1)+C$

$\arcsin (x+2)+C$

$\arctan (x+2)+C$

Correct answer:

$\arctan (x+2)+C$

Explanation:

The integral is found by recognizing the following integral:

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

To solve the integral, make the following substitution:

u=x+2 , du=dx

The derivative was found using the following rule:

$\frac{d}{dx}(x^n)=nx^{n-1}$

The integral then becomes

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

Finish by replacing u with the original term containing x.

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Example Question #12 : Solving Integrals By Substitution

Evaluate the following integral:

$\int{sin(x)cos(x)dx}$

Possible Answers:

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

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

$\frac{cos^2(x)}{2}+c$

$\frac{-sin^2(x)}{2}+c$

Correct answer:

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

Explanation:

Substitution can be used to make this problem easier. Let u = sin(x). Then du = cos(x)dx. This allows us to rewrite and evaluate the original integral in terms of u as:

$\int{udu}=\frac{u^2}{2}+c$

The final answer should be written in terms of the original variables, so substitute sin(x) for u.

$\frac{u^2}{2}+c=\frac{sin^2(x)}{2}+c$

Note that we could have also chosen cos(x) as u, but the above substitution avoids introducing negatives.

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Example Question #13 : Solving Integrals By Substitution

Evaluate the integral $\int{\frac{1}{1+e^{2x}}dx}$ .

Possible Answers:

$e^{2x}+c$

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

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

arctan(e^x)+c

Correct answer:

arctan(e^x)+c

Explanation:

First, notice that $e^{2x}=(e^{x})^2$ . Use the substitution u=e^x, du=e^xdx to rewrite the integral as

$\int{\frac{1}{1+e^{2x}}dx}=\int{\frac{1}{1+u^{2}}du}$ .

Next, recall that $\int\frac{1}{1+x^2}dx=arctan(x)+c$ , so

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

Lastly, substitute e^x in place of to write the answer in terms of the original variables.

arctan(u)+c=arctan(e^x)+c

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Example Question #591 : Finding Integrals

Solve the following indefinite integral:

$\int \cos^4(x)\sin(x)dx$

Possible Answers:

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

$\sec^2(x)+C$

$-\frac{\cos^5(x)}{5}+C$

$\sin^2(x)+C$

Correct answer:

$-\frac{\cos^5(x)}{5}+C$

Explanation:

To solve the indefinite integral, make a simple substitution:

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

The integral then becomes:

$\int -u^4du$

After integrating, we get

$-\frac{u^5}{5}+C$

The following rule was used for the integration:

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

Finally, replace the u with the original term containing x.

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Example Question #14 : Solving Integrals By Substitution

Please solve the following integral:

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

Possible Answers:

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

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

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

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

$\frac{-2}{x(ln(x))^{3}} + C$

Correct answer:

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

Explanation:

We know that the derivative of ln(x) is $\frac{1}x{}$ .

Doing a substitution and setting

$u = \ln x$ and $du = \frac{dx}{x}$ allows us to rewrite the integral as

$\int \frac{du}{u^{2}}$ which can be rewritten as $\int u^{-2}du$ .

Integrating this gets you $\frac{-1}{u}$ plus a constant (which is stated in the original question that you can assume that we already have one). Substituting back in gives us the final answer, which is

$\frac{-1}{ln(x)} + C$ .

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Example Question #11 : Solving Integrals By Substitution

Please solve the following integral.

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

Possible Answers:

$\arctan x + C$

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

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

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

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

Correct answer:

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

Explanation:

We know that the derivative of $\arctan(x)$ is $\frac{1}{1 + x^2}$ .

So substituting

$u = \arctan(x)$ allows us to have

$du= \frac{1}{1 + x^2}$ .

This allows us to rewrite the integral as

$\int u du$ which, when integrated, gives us

$\frac{u^2}{2} + C$ .

Substiting x back in gives us the answer,

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

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Calculus 2 : Integrals

Study concepts, example questions & explanations for Calculus 2

All Calculus 2 Resources

Example Questions

Example Question #2 : Solving Integrals By Substitution

Example Question #4 : Solving Integrals By Substitution

Example Question #9 : Solving Integrals By Substitution

Example Question #3 : Solving Integrals By Substitution

Example Question #11 : Solving Integrals By Substitution

Example Question #12 : Solving Integrals By Substitution

Example Question #13 : Solving Integrals By Substitution

Example Question #591 : Finding Integrals

Example Question #14 : Solving Integrals By Substitution

Example Question #11 : Solving Integrals By Substitution

All Calculus 2 Resources

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