   Chapter 7.1, Problem 50E

Chapter
Section
Textbook Problem

# Prove that, for even powers of sine, ∫ 0 π / 2 sin 2 n x   d x = 1 ⋅ 3 ⋅ 5 ⋅ ⋅ ⋅ ⋅ ⋅ ( 2 n − 1 ) 2 ⋅ 4 ⋅ 6 ⋅ ⋅ ⋅ ⋅ ⋅ 2 n π 2

To determine

To show: the integral 0π2sin2nxdx and 135(2n1)2462nπ2 are equivalent using results from part

Explanation

Reduction formula is obtained by using the technique of integration by parts to solve the integral sinnxdx.

Formula used:

Reduction formula is as given below:

sinnxdx=1ncosxsinn1x+n1nsinn2xdx

Given:

The equation to prove, 0π2sin2nxdx=135(2n1)2462nπ2.

Calculation:

Apply the limits of integration in the reduction formula:

0π2sinnxdx=1ncosxsinn1x]0π2+n1n0π2sinn2xdx=1ncosπ2sinn1π2(1ncos0sinn10)+n1n0π2sinn2xdx

Recall that cosπ2=0 and sin0=0. So, the above integral will become:

0π2sinnxdx=1n(0)sinn1π2(1ncos0(0))+n1n0π2sinn2xdx=n1n0π2sinn2xdx

Solve the integral 0π2sin2nxdx using the formula obtained above with n as 2n. So, substitute for n as 2n in the formula above

0π2sin2nxdx=2n12n0π2sin2n2xdx …… (1)

Solve the integral 0π2sin2n2xdx again using the same formula above with n as 2n2

0π2sin2n2xdx=2n212n20π2sin2n22xdx=2n32n20π2sin2n4xdx

Substitute the result for integration in equation (1)

0

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