Call Now to Set Up Tutoring:

(888) 888-0446

Calculus 2 : Integrals

Study concepts, example questions & explanations for Calculus 2

Create An Account

All Calculus 2 Resources

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

Example Questions

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:

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

Report an Error

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

Report an Error

Example Question #39 : Solving Integrals By Substitution

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

Possible Answers:

-1(x+5)

ln(|x+5|)+C

x+5

ln(x)+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 .

Report an Error

Example Question #971 : Integrals

Possible Answers:

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

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

(x^2-3)^5+C

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

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

Correct answer:

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

Explanation:

Report an Error

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

Report an Error

Example Question #42 : Solving Integrals By Substitution

Evaluate the following indefinite integral:

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

Possible Answers:

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

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

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

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

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,

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

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

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

Report an Error

Example Question #43 : 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

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

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

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

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

Report an Error

Example Question #44 : Solving Integrals By Substitution

Evaluate the following indefinite integral:

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

Possible Answers:

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

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

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

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

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:

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

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

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

Report an Error

Example Question #45 : Solving Integrals By Substitution

Find the value of

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

Possible Answers:

$-\frac{1}{2}$

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

$\ln(2)$

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,

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

Report an Error

Example Question #46 : Solving Integrals By Substitution

Evaluate the following integral:

$\int \frac{dx}{3x\sqrt{9x^2-16}}$

Possible Answers:

$\sec^{-1}(\frac{3x}{4})+C$

$\frac{1}{12}\sec^{-1}(\frac{3x}{4})+C$

$\frac{1}{4}\sec^{-1}(\frac{3x}{4})+C$

$\frac{1}{4}\arctan(\frac{3x}{4})+C$

$-\frac{1}{4}\sec^{-1}(\frac{3x}{4})+C$

Correct answer:

$\frac{1}{12}\sec^{-1}(\frac{3x}{4})+C$

Explanation:

To integrate, we must first make the following substitution:

u=3x, du=3dx

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

$\frac{1}{3}\int \frac{du}{u\sqrt{u^2-16}}=\frac{1}{3}(\frac{1}{4}\sec^{-1}(\frac{u}{4}))+C$

The integral was performed using the following rule:

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

Note that the rule contains a fraction in front of the inverse trig function. Do not confuse this fraction with the fraction coming from the u substitution!

Finally, replace u with our original term and multiply the constants:

$\frac{1}{12}\sec^{-1}(\frac{3x}{4})+C$

Report an Error

View Calculus 2 Tutors

Chandana
Certified Tutor

University of Colombo, Bachelor of Science, Mathematics. University of Wyoming, Doctor of Philosophy, Mathematics.

View Calculus 2 Tutors

Alexander
Certified Tutor

University of Houston, Bachelor of Science, Biomedical Engineering.

View Calculus 2 Tutors

Rodolfo
Certified Tutor

The University of Texas at San Antonio, Bachelor of Science, Mathematics.

All Calculus 2 Resources

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

Popular Subjects

English Tutors in Atlanta, Chemistry Tutors in Boston, MCAT Tutors in Miami, Reading Tutors in Chicago, Math Tutors in Los Angeles, French Tutors in Chicago, Statistics Tutors in Boston, Algebra Tutors in Los Angeles, GRE Tutors in Chicago, SSAT Tutors in Atlanta

Popular Courses & Classes

SSAT Courses & Classes in Dallas Fort Worth, ACT Courses & Classes in Houston, SAT Courses & Classes in Houston, ACT Courses & Classes in Chicago, SAT Courses & Classes in Miami, ISEE Courses & Classes in Houston, LSAT Courses & Classes in Boston, ACT Courses & Classes in Miami, SAT Courses & Classes in Boston, SAT Courses & Classes in San Diego

Popular Test Prep

ACT Test Prep in Chicago, MCAT Test Prep in Phoenix, ACT Test Prep in Houston, GMAT Test Prep in Boston, GRE Test Prep in Los Angeles, MCAT Test Prep in Washington DC, ISEE Test Prep in Dallas Fort Worth, SSAT Test Prep in Houston, LSAT Test Prep in Denver, ACT Test Prep in Denver

DMCA Complaint

If you believe that content available by means of the Website (as defined in our Terms of Service) infringes one or more of your copyrights, please notify us by providing a written notice (“Infringement Notice”) containing the information described below to the designated agent listed below. If Varsity Tutors takes action in response to an Infringement Notice, it will make a good faith attempt to contact the party that made such content available by means of the most recent email address, if any, provided by such party to Varsity Tutors.

Your Infringement Notice may be forwarded to the party that made the content available or to third parties such as ChillingEffects.org.

Please be advised that you will be liable for damages (including costs and attorneys’ fees) if you materially misrepresent that a product or activity is infringing your copyrights. Thus, if you are not sure content located on or linked-to by the Website infringes your copyright, you should consider first contacting an attorney.

Please follow these steps to file a notice:

You must include the following:

A physical or electronic signature of the copyright owner or a person authorized to act on their behalf; An identification of the copyright claimed to have been infringed; A description of the nature and exact location of the content that you claim to infringe your copyright, in \ sufficient detail to permit Varsity Tutors to find and positively identify that content; for example we require a link to the specific question (not just the name of the question) that contains the content and a description of which specific portion of the question – an image, a link, the text, etc – your complaint refers to; Your name, address, telephone number and email address; and A statement by you: (a) that you believe in good faith that the use of the content that you claim to infringe your copyright is not authorized by law, or by the copyright owner or such owner’s agent; (b) that all of the information contained in your Infringement Notice is accurate, and (c) under penalty of perjury, that you are either the copyright owner or a person authorized to act on their behalf.

Send your complaint to our designated agent at:

Charles Cohn Varsity Tutors LLC
101 S. Hanley Rd, Suite 300
St. Louis, MO 63105

Or fill out the form below:

Do not fill in this field

Contact Information

* Full legal name

Company name

* Phone number

* Email address

Address

* Address1

Address2

* City

* State

* Zipcode

* Country

Complaint Details

* URL of copyrighted work (where your original material is located)

* Please describe the copyrighted work so that it may be easily identified

I am the owner, or an agent authorized to act on behalf of the owner of an exclusive right that is allegedly infringed.

I have a good faith belief that the use of the material in the manner complained of is not authorized by the copyright owner, its agent, or the law.

This notification is accurate.

I acknowledge that there may be adverse legal consequences for making false or bad faith allegations of copyright infringement by using this process.

* Signature (your digital signature is legally binding)

HIGH SCHOOL

GRADUATE SCHOOL

K-8

Search 50+ Tests

math tutoring

science tutoring

foreign languages

elementary tutoring

other

Search 350+ Subjects

Calculus 2 : Integrals

Study concepts, example questions & explanations for Calculus 2

All Calculus 2 Resources

Example Questions

Example Question #37 : Solving Integrals By Substitution

Example Question #38 : Solving Integrals By Substitution

Example Question #39 : Solving Integrals By Substitution

Example Question #971 : Integrals

Example Question #41 : Solving Integrals By Substitution

Example Question #42 : Solving Integrals By Substitution

Example Question #43 : Solving Integrals By Substitution

Example Question #44 : Solving Integrals By Substitution

Example Question #45 : Solving Integrals By Substitution

Example Question #46 : Solving Integrals By Substitution

All Calculus 2 Resources

Report an issue with this question

DMCA Complaint

Contact Information

Address

Complaint Details

Email address:
Your name:
Feedback: