   Chapter 7.2, Problem 3E

Chapter
Section
Textbook Problem

# Evaluate the integral. ∫ 0 π / 2 sin 7 θ   cos 5 θ   d θ

To determine

To evaluate: The trigonometric integral 0π2sin7θcos5θdθ

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 both sine and cosine in the integral is odd, save one sine (or cosine) factor and use the identity sin2x=1cos2x to rewrite other terms in cosine (or sine) function form. Here, lets save one sine function and convert rest to cosine terms.

sin2k+1xcosnxdx=cosnx(1cos2x)ksinxdx

Then, use the substitution u=cosx

Given:

The integral, 0π2sin7θcos5θdθ.

Calculation:

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

0π2sin7θcos5θdθ=0π2sin6θcos5θsinθdθ

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

0π2sin6θcos5θsinθdθ=0π2(1cos2θ)3cos5θsinθdθ

Use the substitution u=cosθ and du=sinθdθ. So, the integral will be:

0π2(1cos2θ)3cos5θsinθdθ=0π2(1u2)3u5(du)=0π2(1u2)3u5du=0π2(13312u2+31(u2)2(u2)3)u5<

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