   Chapter 7.P, Problem 9P

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# Show that ∫ 0 1 ( 1 − x 2 ) n d x = 2 2 n ( n ! ) 2 ( 2 n + 1 ) ! Hint: Start by showing that if I n denotes the integral, then I k + 1 = 2 k + 2 2 k + 3 I k

To determine

To Prove:. In=01(1x2)ndx=22n(n!)2(2n+1)!

Explanation

Given: The expressionis 01(1x2)ndx

Concepts used: Mathematical Induction, Trigonometric substitution.

Calculation::

Consider 01(1x2)ndx

We shall prove the given statement using induction and trigonometric substitution.For the case n=1

We have 01(1x2)dx=[xx33]01=23=46=22(1!)2(2+1)!

Hence the statement is true for n =1. Suppose it is true for n=k for some positive integer k.That is Ik=01(1x2)kdx=22k(k!)2(2k+1)! Now consider Ik+1=01(1x2)k+1dx Let

x=sinθ then dx=cosθdθ since  θ=sin1x,when x=0, θ=0 and when x=1, θ=π/2

Thus

01(1x2)k+1dx=0π/2(1sin2θ)k+1cosθdθ=0π/2(cos2θ)k+1cosθdθ=0π/2cos2k+3θdθ                               (1)

Recall the reduction formula for cosine (it is entry number 74 in integral table) It says cosnθdθ=1ncosn1θsinθ+n1ncosn2θdθ Applying this to (1) we have 0π/2cos2k+3θdθ  = [12k+3cos2k+2θsinθ]0π/2+2k+22k+30π/2cos2k+1θdθcosnπ2=0, for all n>0 andsin0=0

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