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Let $(I_n)_{n \geq 1}$ be a sequence such that: $$I_n = \int_0^1 (1-x^2)^n dx$$

Find the value of $I_n$ (The solution is $\frac{2}{3} \cdot \frac{4}{5} \cdot ... \cdot \frac{2n}{2n+1}$).

I've tried using the binomial expansion of $(1-x^2)^n$ but I couldn't get the proper answer.

Thank you!

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    Put $x=\sin y$ and use http://calc101.com/deriving_reduction_2.html – 2017-02-26

1 Answers 1

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I've tried using the binomial expansion of $(1-x^2)^n$ but I couldn't get the proper answer.

One may use integration by parts obtaining $$ \begin{align} I_n=\int_{0}^{1}(1-x^2)^ndx&=\left[x(1-x^2)^n\right]_{0}^{1}+2n\int_{0}^{1}x^2(1-x^2)^{n-1}dx \\\\&=0+2n\int_{0}^{1}\left[(1-(1-x^2))(1-x^2)^{n-1}\right]dx \\\\&=2nI_{n-1}-2nI_{n} \end{align} $$ then, using $I_0=1$, $I_1=\frac23,$ $$ I_{n}=\frac{2n}{2n+1}\cdot I_{n-1}, \quad n\ge1, $$ one gets the announced result:

$$I=\int_{0}^{1}(1-x^2)^ndx=\frac{2}{3} \cdot \frac{4}{5} \cdots \frac{2n}{2n+1}={(2n)!!\over (2n+1)!!}, \qquad n\ge1.$$