Call Now to Set Up Tutoring:

(888) 888-0446

High School Math : Finding Integrals

Study concepts, example questions & explanations for High School Math

Create An Account

All High School Math Resources

8 Diagnostic Tests 613 Practice Tests Question of the Day Flashcards Learn by Concept

Example Questions

Example Question #61 : Finding Integrals

Determine the indefinite integral:

$\dpi{120} \int\cos x \sqrt{\sin x }\; \; dx$

Possible Answers:

$\dpi{120} \dpi{120} \frac{ 2 \sqrt{\cos x }} {3} +C$

$\dpi{120} \frac{ 2 \sin x \cdot \sqrt{\cos x }} {3} +C$

$\dpi{120} \dpi{120} \frac{ 2 \cos x \cdot \sqrt{\cos x }} {3} +C$

$\dpi{120} \frac{ 2 \sqrt{\sin x }} {3} +C$

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

Correct answer:

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

Explanation:

Set $u = \sin x$ . Then

$\frac{\mathrm{du} }{\mathrm{d} x}= \frac{\mathrm{d} }{\mathrm{d} x} \sin x= \cos x$ .

and

$\cos x \; dx = du$

The integral becomes:

$\dpi{120} \int\cos x \sqrt{\sin x }\; \; dx$

$\dpi{120} =\int \sqrt{u }\; \; du$

$= \int u ^{\frac{1}{2}} \; du$

$\dpi{120} \dpi{150} = \frac{u^{\frac{1}{2}+1}}{{\frac{1}{2}+1}} + C$

$\dpi{150} = \frac{u^{\frac{3}{2}}}{{\frac{3}{2}}} + C$

$\dpi{120} =\frac{ 2u \sqrt{u}} {3} +C$

Substitute back:

$\dpi{120} =\frac{ 2 \sin x \cdot \sqrt{\sin x }} {3} +C$

Report an Error

Example Question #4 : Finding Integrals By Substitution

$\int_{4}^{9}\frac{\sqrt{x}}{1-x}dx=$

Possible Answers:

$-2+\ln\frac{9}{4}$

$-2+\ln\frac{3}{2}$

$-2+\ln\frac{2}{3}$

$\tan^{-1}(3)-\tan^{-1}(4)$

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

Correct answer:

$-2+\ln\frac{2}{3}$

Explanation:

This integral will require a u-substitution.

Let $u=\sqrt{x}$ .

Then, differentiating both sides, $du=\frac{1}{2\sqrt{x}}dx$ .

We need to solve for dx in order to replace all x terms with u terms.

$dx=(2\sqrt{x})du$ .

$\int\frac{\sqrt{x}}{1-x}\cdot(2\sqrt{x})du=\int\frac{2xdu}{1-x}$

This is a little tricky because we stilll have x and u terms mixed together. We need to go back to our original substitution.

$u=\sqrt{x}\Rightarrow x=u^2$

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

Now we have an integral that looks more manageable. First, however, we can't forget about the bounds of the definite integral. We were asked to evaluate the integral from x=4 to x=9 . Because $u=\sqrt{x}$ , the bounds will change to $\sqrt{4}=2$ and $\sqrt{9}=3$ .

Essentially, we have made the following transformation:

$\int_{4}^{9}\frac{\sqrt{x}}{1-x}dx=\int_{2}^{3}\frac{2u^2}{1-u^2}du$ .

The latter integral is easier to evaluate.

$\int_{2}^{3}\frac{2u^2}{1-u^2}du=-2\int_{2}^{3}\frac{-u^2}{1-u^2}du$

$=-2\int_{2}^{3}\frac{1-u^2-1}{1-u^2}du=-2\int_{2}^{3}\left ( \frac{1-u^2}{1-u^2}+\frac{-1}{1-u^2} \right )du$

$=-2\int_{2}^{3}\left ( 1-\frac{1}{1-u^2}\right )du$

At this point, we can separate the integral into two smaller integrals.

$-2\int_{2}^{3}\left ( 1-\frac{1}{1-u^2}\right )du=-2\int_{2}^{3}1\cdot du+2\int_{2}^{3}\frac{1}{1-u^2}du$ .

The integral $-2\int_{2}^{3}1\cdot du$ evaluates to -2, so now we just need to worry about the other integral. This will require the use of partial fraction decomposition. We need to rewrite $\frac{1}{1-u^2}$ as the sum of two fractions.

$\frac{1}{1-u^2}=\frac{1}{(1-u)(1+u)}=\frac{A}{1-u}+\frac{B}{1+u}$

We need to solve for the values of A and B.

$\frac{A}{1-u}+\frac{B}{1+u}=\frac{1}{(1-u)(1+u)}$

A(1+u)+B(1-u)=1=0u+1

A+Au+B-Bu=0u+1

Au-Bu+A+B=0u+1

(A-B)u+(A+B)=0u+1

This means that A-B=0 and A+B=1 . This is a relatively simple system of equations to solve, so I won't go into detail. The end result is that $A=B=\frac{1}2{}$ .

$\frac{1}{1-u^2}=\frac{\frac{1}{2}}{1-u}+\frac{\frac{1}{2}}{1+u}$

Let's now go back to the integral $2\int_{2}^{3}\frac{1}{1-u^2}du$ .

$2\int_{2}^{3}\frac{1}{1-u^2}du=2\int_{2}^{3}\left ( \frac{\frac{1}{2}}{1-u}+\frac{\frac{1}{2}}{1+u} \right )du$

Distribute the 2 to both integrals and separate it into two integrals.

$=\int_{2}^{3}\frac{1}{1-u}du + \int_{2}^{3}\frac{1}{1+u}du$

$=-\ln|1-u| ]_{2}^{3} + \ln|1+u|]_{2}^{3}$

$=-\ln|-2|+\ln|-1|+\ln4-\ln3$

$=-\ln2+\ln1+\ln4-\ln3$

$=\ln\frac{2}{3}$ .

Remember we need to add this value back to the value of $-2\int_{2}^{3}1\cdot du$ , which we already determined to be -2.

The final answer is $-2+\ln\frac{2}{3}$ .

Report an Error

Example Question #1 : Finding Integrals By Substitution

Evaluate:

$\dpi{120} \int_{0}^{\frac{\pi }{ 2}}\sin x \sqrt{\cos x }\; \; dx$

Possible Answers:

$\frac{ 2} {3}$

$\frac{4}{3}$

Correct answer:

$\frac{ 2} {3}$

Explanation:

Set $u = \cos x$ .

Then $\frac{\mathrm{du} }{\mathrm{d} x}= \frac{\mathrm{d} }{\mathrm{d} x} \cos x= -\sin x$ and $\sin x \; dx = - du$ .

Also, since $u = \cos x$ , the limits of integration change to $\cos \frac{\pi }{2} = 0$ and $\cos 0 = 1$ .

Substitute:

$\dpi{120} \int_{0}^{\frac{\pi }{ 2}}\sin x \sqrt{\cos x }\; \; dx$

$= \int_{1}^{0} \left (- \sqrt{u} \right ) \; du$

$= \int_{1}^{0} \left (-u ^{\frac{1}{2}} \right ) \; du$

$= \int_{0}^{1} u ^{\frac{1}{2}} \; du$

$\dpi{150} = \left.\begin{matrix} \frac{u^{\frac{3}{2}}}{{\frac{3}{2}}} & \\ & \end{matrix}\right|$ $\begin{matrix} 1\\ \\ 0 \end{matrix}$

$\dpi{150} = \left.\begin{matrix} \\ \frac{ 2u \sqrt{u}} {3} & \\ & \end{matrix}\right|$ $\begin{matrix} 1\\ \\ 0 \end{matrix}$

$=\frac{ 2\cdot 1\cdot \sqrt{1}} {3} - \frac{ 2\cdot 0\cdot \sqrt{0}} {3} = \frac{ 2} {3}$

Report an Error

View Tutors

Rebekah
Certified Tutor

Swarthmore College, Bachelor, Psychology & Linguistics. Pace University, Master's/Graduate, School Psychology.

View Tutors

Joy
Certified Tutor

Boston University, Bachelor in Arts, Conservation Biology.

View Tutors

Maryanne Federici
Certified Tutor

Brown University, Bachelor in Arts, Biology, General. The University of Texas at Austin, Master of Arts, Music.

All High School Math Resources

8 Diagnostic Tests 613 Practice Tests Question of the Day Flashcards Learn by Concept

Popular Subjects

Algebra Tutors in Los Angeles, GRE Tutors in Chicago, Physics Tutors in San Francisco-Bay Area, SSAT Tutors in Boston, English Tutors in Denver, Statistics Tutors in Phoenix, Reading Tutors in San Francisco-Bay Area, English Tutors in Atlanta, Biology Tutors in Chicago, Chemistry Tutors in New York City

Popular Courses & Classes

ISEE Courses & Classes in Philadelphia, ISEE Courses & Classes in Houston, MCAT Courses & Classes in Phoenix, ACT Courses & Classes in Philadelphia, SAT Courses & Classes in Washington DC, MCAT Courses & Classes in Boston, Spanish Courses & Classes in San Francisco-Bay Area, GMAT Courses & Classes in Los Angeles, SAT Courses & Classes in Denver, Spanish Courses & Classes in Chicago

Popular Test Prep

GRE Test Prep in Phoenix, ISEE Test Prep in San Diego, MCAT Test Prep in Miami, MCAT Test Prep in Chicago, GMAT Test Prep in Los Angeles, LSAT Test Prep in Denver, LSAT Test Prep in Washington DC, GMAT Test Prep in Philadelphia, GMAT Test Prep in Dallas Fort Worth, ISEE Test Prep in Miami

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

High School Math : Finding Integrals

Study concepts, example questions & explanations for High School Math

All High School Math Resources

Example Questions

Example Question #61 : Finding Integrals

Example Question #4 : Finding Integrals By Substitution

Example Question #1 : Finding Integrals By Substitution

All High School Math Resources

Report an issue with this question

DMCA Complaint

Contact Information

Address

Complaint Details

Email address:
Your name:
Feedback: