   Chapter 7.2, Problem 4E

Chapter
Section
Textbook Problem

# Evaluate the integral. ∫ 0 π / 2 sin 5 x   d x

To determine

To evaluate: The trigonometric integral 0π2sin5xdx

Explanation

Trigonometric integral of the form sinmxcosnxdx can be solved using strategies depending on whether m and n are odd or even.

Formula used:

When power of sine in the integral is odd, save one sine factor and use the identity sin2x=1cos2x to rewrite other terms in cosine function form:

sin2k+1xcosnxdx=cosnx(1cos2x)ksinxdx

Then, use the substitution u=cosx

Given:

The integral, 0π2sin5xdx.

Calculation:

Rewrite the given integral in terms of cosine, saving one sine term:

0π2sin5xdx=0π2sin4xsinxdx

Use the identity sin2x+cos2x=1, to convert remaining terms into cosine:

0π2sin4xsinxdx=0π2(sin2x)2sinxdx=0π2(1cos2x)2sinxdx

Use the substitution u=cosx and du=sinxdx

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