   Chapter 4.5, Problem 66E

Chapter
Section
Textbook Problem

# Use Exercise 65 to evaluate ∫ 0 π / 2 cos 2 x d x and ∫ 0 π / 2 sin 2 x d x .

To determine

To evaluate:

0π2cos2xdx and 0π2sin2xdx

Explanation

1) Concept:

i. The substitution rule: If u=g(x) is a differentiable function whose range is I and f is continuous on I, then f(gx)g'xdx=f(u)du. Here g(x) is substituted as u and then differentiation g(x)dx =du

ii.

abf(x)dx=-baf(x)dx

iii.

abf(u)du=abf(x)dx

2) Formula:

i.

cosπ2-x=sinx

ii. sin2x+cos2x=1

3) Calculation:

Using exercise 65,

0π2f(cosx)dx=0π2f(sinx)dx

Let fx=x2

So, from above we have

0π2cos2xdx=0π2sin2xdx

Now add 0π2cos2xdx on both sides

0π2cos2xdx+0π2cos2xdx=0π2sin2xdx+0π2cos2xdx

Simplify

2

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