   Chapter 7.2, Problem 6E

Chapter
Section
Textbook Problem

# Evaluate the integral. ∫ t   cos 5 ( t 2 ) d t

To determine

To evaluate: The trigonometric integral tcos5(t2)dt

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

sinmxcos2k+1xdx=sinmx(1sin2x)kcosxdx

Then, use the substitution u=sinx

Given:

The integral, tcos5(t2)dt.

Calculation:

First make the substitution t2=θ, which gives 2tdt=dθ. Then, the integral will become

tcos5(t2)dt=tcos5(θ)dθ2t=12cos5(θ)dθ

Now, rewrite the integral in terms of sine, saving one cosine term:

12cos5(θ)dθ=12cos4θcosθdθ

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

12cos4θcosθdθ=12(cos2θ)2cosθdθ=12(1sin2θ)2cosθdθ

Use the substitution u=sinθ and du=cosθdθ

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