Prove that : $$\int_{-1}^{1} (1-x^2)^n dx = \frac{(n!)^2 2^{2n + 1}}{(2n+1)!}$$
Prove that $\int_{-1}^{1} (1-x^2)^n dx = \frac{(n!)^2 2^{2n + 1}}{(2n+1)!}$
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integration
definite-integrals
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1Any own thoughts or tries? What have you tried? – 2017-01-15
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0i would say induction – 2017-01-15
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0Observe that $\int_{-1}^1(1-x^2)^n\,dx=2\int_0^1(1-x^2)^n\,dx$, then use a change of variable : $x=\sin(\theta)$. This should lead to a Wallis integral. – 2017-01-15
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0What I try is to use binomial theorem for $(1-x^2)^n$ and then integrate . – 2017-01-15
2 Answers
0
One way: $$\int_{-1}^{1} (1-x^2)^n dx\underbrace{=}_{f\text{ even}}2\int_{0}^{1} (1-x^2)^ndx\underbrace{=}_{t=x^2}\int_{0}^{1} (1-t)^nt^{-1/2}dt=B(n+1,1/2)=\ldots$$
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The trick is induction together with integration by parts. Denote our integral by $I_n$. Then
$$ I_{n+1}=\int_{-1}^1(1-x^2)^n(1-x^2)\text{d}x=I_n-\int_{-1}^1 x^2(1-x^2)^n\text{d}x. $$
Integrating by parts by $u=x$ and $v'=x(1-x^2)^n$ we arrive at the recurrence $$I_{n+1}=\frac{2n+2}{2n+3}I_n.$$
Therefore $$I_{n+1}=\frac{2(n+1)\cdot 2(n+1)(n!)^2\cdot 2^{2n+1}}{2(2n+3)(n+1)(2n+1)!}=\frac{\bigl((n+1)!\bigr)^2\cdot 2^{2n+3}}{(2n+3)!},$$ which is the induction step. I leave to you the verification of the formula for $n=1$.