   Chapter 5.5, Problem 58E ### Single Variable Calculus: Early Tr...

8th Edition
James Stewart
ISBN: 9781305270343

#### Solutions

Chapter
Section ### Single Variable Calculus: Early Tr...

8th Edition
James Stewart
ISBN: 9781305270343
Textbook Problem

# Evaluate the definite integral. ∫ π / 3 2 π / 3 csc 2 ( 1 2 t )   d t

To determine

To evaluate: The definite integral.

Explanation

Given:

The definite integral function is π32π3csc2(12t)dt.

The region lies between t=π3 and t=2π3.

Calculation:

Consider u=12t (1)

Differentiate both sides of the Equation (1).

du=12dtdt=2du

Calculate the lower limit value of u using Equation (1).

Substitute π3 for t in Equation (1).

u=12(π3)=π6

Calculate the upper limit value of u using Equation (1).

Substitute 2π3 for t in Equation (1).

u=12(2π3)=2π6

The definite integral function is,

π32π3csc2(12t)dt (2)

Apply lower and upper limits for u in Equation (2).

Substitute u for (12t) and (2du) for dt in Equation (2).

π32π3csc2(12t)dt=π62π6csc2u(2du)=2π62π6csc2udu (3)

Integrate Equation (3)

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