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Calculus 2 : Finding Integrals

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9 Diagnostic Tests 308 Practice Tests Question of the Day Flashcards Learn by Concept

Example Questions

Example Question #36 : Solving Integrals By Substitution

Evaluate $\int^{\pi/2}_{0}\frac{2x\tan^{-1}(x^2)}{1+x^4}dx$

Possible Answers:

$\frac{\pi}{2}$

$\frac{\pi}{3}$

None of the other answers.

Correct answer:

None of the other answers.

Explanation:

The correct answer is $\frac{1}{2}(\tan^{-1}(\frac{\pi^2}{4}))^2$ .

We proceed as follows-

$\int^{\pi/2}_{0}\frac{2x\tan^{-1}(x^2)}{1+x^4}dx$ . Start

Evaluating this integral relies on the fact $\frac{d}{dx}\tan^{-1}(x) = \frac{1}{1+x^2}$ , and the Chain Rule for derivatives.

Use u-substitution $u = \tan^{-1}(x^2)$ , then we obtain $du = \frac{1}{1+(x^2)^2} \times 2x dx$

Our integral then becomes

$\int^{\tan^{-1}(\frac{\pi^2}{4})}_{0} u du$ after substitution. (The new upper bound on the integral cannot be simplified well, so we should leave it as is).

We then integrate to get

$[\frac{1}{2}u^2]^{\tan^{-1}(\frac{\pi^2}{4})}_0$

$\frac{1}{2}(\tan^{-1}(\frac{\pi^2}{4}))^2$

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Example Question #37 : Solving Integrals By Substitution

Evaluate the following integral:

$\int \frac{dx}{36+4x^2}$

Possible Answers:

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

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

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

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

Correct answer:

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

Explanation:

To integrate, we must first make the following substitution:

u=2x, du=2dx

Next, rewrite the integral and integrate:

$\frac{1}{2}\int \frac{du}{36+u^2}=\frac{1}{2}\arctan(\frac{u}{6})+C$

The integration was performed using the following rule:

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

Finally, replace u with our original x term:

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

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Example Question #38 : Solving Integrals By Substitution

What is the integral of the following equation?

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

Possible Answers:

$y=5\sqrt{x^{2}+1}+C$

$y=5sin^{-1}x+C$

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

$y=ln(\sqrt{x^{2}+1})+C$

Correct answer:

$y=5\sqrt{x^{2}+1}+C$

Explanation:

We can solve this integral with u substitution

$y=\int \frac{5x}{\sqrt{x^{2}+1}} dx$ let $u=x^{2}+1$ , so du=2xdx , or, $\frac{1}{2}du=xdx$

Making this substitution, and moving our constants gives us:

$y=\frac{5}{2}\int \frac{du}{\sqrt{u}}$ , solving the integral, we get $y=\frac{5}{2}(2)\sqrt{u}+C$ , plugging our value for u back into the equation $y=5\sqrt{x^{2}+1}+C$

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Example Question #971 : Integrals

$\int \frac{1}{x+5} \; dx$

Possible Answers:

-1(x+5)

ln(x)+C

x+5

ln(|x+5|)+C

$-1(x+5)^{-2}$

Correct answer:

ln(|x+5|)+C

Explanation:

To make this integral simpler, we will need to make a substitution. You want to pick a substitution where the derivative also exists in the integral. Here, we want to choose:

u=x+5,du=1dx . Now, we want to rewrite the integral interms of the new variable.

$\int \frac{1}{u} \;du=ln(|u|)+C$ .

The last step is just to substitute the original substitution back in.

ln(|u|)+C=ln(|x+5|)+C .

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Example Question #621 : Finding Integrals

Possible Answers:

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

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

$\frac{(x^2-3)^5}{5}+C$

$\frac{(x^2-3)^5}{5}$

(x^2-3)^5+C

Correct answer:

$\frac{(x^2-3)^5}{5}+C$

Explanation:

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Example Question #41 : Solving Integrals By Substitution

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

Solve the indefinite integral using trigonemtric substitution

Possible Answers:

arcsec(3x)+c

arctan(3x)+c

arccos(3x)+c

arcsin(3x)+c

Correct answer:

arcsec(3x)+c

Explanation:

We substitute

$x=\frac{sec(\theta)}{3}$

to solve the integral. Solving for dx,

$dx=\frac{1}{3}sec(\theta)tan(\theta)d\theta$

Substituting these values into the integral yields

$\int \frac{\frac{1}{3}sec(\theta)tan(\theta)}{\frac{sec(\theta)}{3}\sqrt{9(\frac{sec(\theta)}{3})^2-1}}\,d\theta$

$\int \frac{\frac{1}{3}sec(\theta)tan(\theta)}{\frac{sec(\theta)}{3}\sqrt{sec^2(\theta)-1}}\,d\theta$

$\int \frac{\frac{1}{3}sec(\theta)tan(\theta)}{\frac{sec(\theta)}{3}\sqrt{tan^2(\theta)}}\,d\theta$

$\int \frac{\frac{1}{3}sec(\theta)tan(\theta)}{\frac{1}{3}sec(\theta)\tan(\theta)}\,d\theta$

$\int 1\,d\theta=\theta+c$

Solving for $\theta$ from

$x=\frac{sec(\theta)}{3}$

gives us

$\theta=arcsec(3x)$

And so the indefinite integral is

arcsec(3x)+c

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Example Question #41 : Solving Integrals By Substitution

Evaluate the following indefinite integral:

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

Possible Answers:

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

$2\ln|x^2+1|+C$

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

$2\ln|x^2-1|+C$

Correct answer:

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

Explanation:

The integrand is composed of a function as well as its derivative multiplied by a constant. Hence, we can find the antiderivative via u-substitution as follows:

Let u = x^2-1 . Then du = 2xdx , and so xdx = 1/2du . Thus,

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

$=\frac{1}{2} \ln|u|+C$

$= \frac{1}{2}\ln|x^2-1|+C$

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Example Question #42 : Solving Integrals By Substitution

Evaluate the following indefinite integral:

$\int \frac{1}{(x+7)^4} dx$

Possible Answers:

$-\frac{1}{5}(x+7)^{-5}+C$

$-\frac{1}{3}(x+7)^-^3+C$

$\frac{1}{3}(x+7)^3+C$

$\frac{1}{5}(x+7)^5+C$

Correct answer:

$-\frac{1}{3}(x+7)^-^3+C$

Explanation:

The integrand can be evaluated by means of the u-substitution method, as follows:

Let u = x+7 . Then du = dx , and so

$\int \frac{1}{(x+7)^4} \ dx = \int \frac{1}{u^4}\ du$

$=\int u^{-4}\ du$

$= -\frac{1}{3}u^{-3} +C$

$=-\frac{1}{3}(x+7)^{-3}+C$

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Example Question #43 : Solving Integrals By Substitution

Evaluate the following indefinite integral:

$\int sin^5x cos^2x dx$

Possible Answers:

$\frac{1}{3}sin^3x-\frac{2}{5}sin^5x+\frac{1}{7}sin^7x+C$

$\frac{1}{3}cos^3x-\frac{2}{5}cos^5x+\frac{1}{7}cos^7x+C$

$-\frac{1}{3}xsin^3x+\frac{2}{5}sin^5x-\frac{1}{7}sin^7x+C$

$-\frac{1}{3}cos^3x+\frac{2}{5}cos^5x-\frac{1}{7}cos^7x+C$

Correct answer:

$-\frac{1}{3}cos^3x+\frac{2}{5}cos^5x-\frac{1}{7}cos^7x+C$

Explanation:

Here, an understanding of trigonometric identities, as well as the appropriate selection of a dummy variable for u-substitution, is required. To figure out which function to represent "u" (cosine or sine), simply re-write the integrand as

$\int sin^5xcos^2x dx = \int sin^4x cos^2x sinx dx$

$=\int (sin^2x)^2cos^2xsinx dx$

Remembering that sin^2x +cos^2x = 1 ,

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

Now, we can substitute u = cosx to yield

$\int (1-cos^2x)^2cos^2xsinxdx = \int (1-u^2)^2u^2(-du)$

because if u = cosx , then du = -sinx dx , which implies -du = sinx dx .

At this point, all that is left to do is expand the polynomial and evaluate the integrand:

$\int (1-u^2)^2u^2(-du) = -\int (u^2-2u^4+u^6) \ du$

$=-(\frac{1}{3}u^3-\frac{2}{5}u^5+\frac{1}{7}u^7)+C$

$=-\frac{1}{3}cos^3(x)+\frac{2}{5}\cos^5(x)-\frac{1}{7}\cos^7(x)+C$

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Example Question #625 : Finding Integrals

Find the value of

$\int_{e}^{e^2}\frac{1}{xln(x)}dx$ .

Possible Answers:

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

$\ln(2)$

$-\frac{1}{2}$

Correct answer:

$\ln(2)$

Explanation:

To perform this integration, we use a substitution.

Since the derivative of $\ln(x)$ is $\frac{1}{x}$ , we choose our substitution to be $u=\ln(x)$ .

Differentiating gives us,

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

Now we can substitute this into our integral. We will have,

$\frac{1}{x\ln(x)}dx=\frac{1}{ln(x)}\frac{1}{x}dx=\frac{1}{u}du$ .

Along with this substitution, we must also change our limits of and e^2 . To do so, we take these values and plug them in for in the formula $u=\ln(x)$ .

Doing so, we obtain $u_{upper}=\ln(e^2)=2$ and $u_{lower}=\ln(e)=1$ .

Now our integral will be transformed as follows,

$\int_{e}^{e^2}\frac{1}{xln(x)}dx=\int_1^2\frac{1}{u}du$ .

This integral is now easy to integrate, for the function $\frac{1}{u}$ integrates to $\ln(u)$ .

Thus we have,

$\int_1^2\frac{1}{u}du=\ln(u)\bigg|_1^2=\ln(2)-\ln(1)=\ln(2)$ .

Therefore, the answer to the integral is,

$\int_{e}^{e^2}\frac{1}{xln(x)}dx=\ln(2)$ .

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Calculus 2 : Finding Integrals

Study concepts, example questions & explanations for Calculus 2

All Calculus 2 Resources

Example Questions

Example Question #36 : Solving Integrals By Substitution

Example Question #37 : Solving Integrals By Substitution

Example Question #38 : Solving Integrals By Substitution

Example Question #971 : Integrals

Example Question #621 : Finding Integrals

Example Question #41 : Solving Integrals By Substitution

Example Question #41 : Solving Integrals By Substitution

Example Question #42 : Solving Integrals By Substitution

Example Question #43 : Solving Integrals By Substitution

Example Question #625 : Finding Integrals

All Calculus 2 Resources

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