   Chapter 7.1, Problem 28E

Chapter
Section
Textbook Problem

# Evaluate the integral. ∫ 0 2 π t 2 sin 2 t   d t

To determine

To evaluate: The given integral using the technique of integration by parts.

Explanation

The technique of integration by parts comes in handy when the integrand involves product of two functions. It can be thought of as a rule corresponding to the product rule in differentiation.

Formula used:

The formula for integration by parts for definite integral is given by

abf(x)g(x)dx=f(x)g(x)]ababg(x)f(x)dx

Given:

The integral, 02πt2sin2tdt.

Calculation:

Make the choice for u and dv such that the resulting integration from the formula above is easier to integrate. Let

u=t2      dv=sin2tdt

Then, the differentiation of u and antiderivative of dv will be

du=2tdt     v=12cos2t

Using the formula above, the given integration will become

02πt2sin2tdt=t2(12cos2t)]02π02π(12cos2t)2tdt=t2(12cos2t)]02π+02πtcos2tdt …… (1)

Integral given by the last term is again a product of two functions. Hence, solve it by using integration by parts with the variables as

u=t     dv=cos2tdt

Then

du=dt      v=12sin2t

Then, the integration 02πtcos2tdt will be

02πtcos2tdt

### Still sussing out bartleby?

Check out a sample textbook solution.

See a sample solution

#### The Solution to Your Study Problems

Bartleby provides explanations to thousands of textbook problems written by our experts, many with advanced degrees!

Get Started

#### Evaluate the integral, if it exists. 01y(y2+1)5dy

Single Variable Calculus: Early Transcendentals, Volume I

#### Rationalize Numerator Rationalize the numerator. 95. x2+1x

Precalculus: Mathematics for Calculus (Standalone Book)

#### Solve the equations in Exercises 126. (x21)2(x+2)3(x21)3(x+2)2=0

Finite Mathematics and Applied Calculus (MindTap Course List)

#### In Problems 5-8, find the derivative but do not simplify your answer. 7.

Mathematical Applications for the Management, Life, and Social Sciences 