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Consider these two integrals

$$\int_{0}^{1}{1-x^{2n+1}\over (1-x^2)^{3/2}}\mathrm dx=I\tag1$$

$$\int_{0}^{1}{1-x^{2n}\over (1-x^2)^{3/2}}\mathrm dx={n\pi\over I}\tag2$$

An attempt: $(1)$

$x=\sin{u}$ then $dx=\cos{u}du$

$(1)$ becomes

$$\int_{0}^{\pi/2}(1-\sin^{2n+1}{u})\cdot{\mathrm du\over \cos^2{u}}\tag3$$

Applying integration by parts to $(3)$ seem long because of evaluating $\sin^{2n+1}{u}$

How else can we evaluate $(1)$ and $(2)$ using another way?

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Both $(1)$ and $(2)$ can be computed in terms of the $\Gamma$ function by exploiting integration by parts, the substitution $x=\sqrt{z}$ and Euler's Beta function. For instance $$ \int_{0}^{1}\frac{1-x^{2n+1}}{(1-x^2)^{3/2}}\,dx = \frac{\sqrt{\pi}\,\,\Gamma(n+1)}{\Gamma\left(n+\frac{1}{2}\right)}\tag{1}$$ $$ \int_{0}^{1}\frac{1-x^{2n}}{(1-x^2)^{3/2}}\,dx = \frac{\sqrt{\pi}\,\,\Gamma\left(n+\frac{1}{2}\right)}{\Gamma\left(n\right)}\tag{2}$$ and the claim $(1)\cdot(2)=\pi\,n$ trivially follows.

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  • $\begingroup$ Hey Jack ,maybe it can be proved without evaluating each one explicitly? $\endgroup$ – Zaid Alyafeai Feb 25 '17 at 22:06
  • $\begingroup$ @ZaidAlyafeai: probably yes, the problem is essentially asking us to simplify $\int_{0}^{\pi/2}\sin(\theta)^{2n}\,d\theta \int_{0}^{\pi/2}\sin(\varphi)^{2n+1}\,d\varphi$ that can be reduced to a simple integral over the unit ball in $\mathbb{R}^3$, by Fubini's theorem. $\endgroup$ – Jack D'Aurizio Feb 25 '17 at 22:13
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    $\begingroup$ $$ \begin{align} \color{red}{I}&=\int_0^1\frac{1-x^{2n+1}}{(1-x^2)^{3/2}}\,dx \\ &\qquad u=1-x^{2n+1}\Rightarrow du=-(2n+1)x^{2n}\quad\&\quad dv=\frac{dx}{(1-x^2)^{3/2}}\Rightarrow v=\frac{x}{\sqrt{1-x^2}} \\ &=(2n+1)\int_0^1x^{2n+1}{(1-x^2)^{-1/2}}\,dx=(n+1/2)\,B(n+1,1/2)=\color{red}{\frac{\sqrt{\pi}\,\Gamma(n+1)}{\Gamma(n+1/2)}} \end{align} $$ Unfortunately I am not as fast as U R, (+1). $\endgroup$ – Hazem Orabi Feb 25 '17 at 23:10
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Well after a long time I managed to come up with a formula for $I_{n}$

$I_{n} = \int_{0}^{\frac{\pi}{2}}\frac{1- \sin^{2n+1}u}{1-\sin^{2}u}du$

$I_{n} = -\int_{0}^{\frac{\pi}{2}}\frac{(\sin u-1)(\sin ^{2n}u + \sin ^{2n-1}u +... +\sin u + 1)}{(1-\sin u)(1+\sin u)}du$

$I_{n} = \int_{0}^{\frac{\pi}{2}}\frac{(\sin u + 1)(\sin^{2n-1}u + \sin^{2n-3}u +...\sin u + 1)}{\sin u +1}du$

$I_{n} = \int_{0}^{\frac{\pi}{2}} \sin^{2n-1}u + ... + \sin u du +\int_{0}^{\frac{\pi}{2}}\frac{1}{\sin u + 1} du$

We can calculate $\int_{0}^{\frac{\pi}{2}}\frac{1}{\sin u + 1} du = 1$

Then $I_{n} = \int_{0}^{\frac{\pi}{2}} \sin^{2n-1}u + ... + \sin u du +1$

Consider $J_{n} = \int_{0}^{\frac{\pi}{2}} \sin^{2n+1}u du$

We can use reduction formula to derive that $J_{n} = \frac{2^{2n+1}n!(n+1)!}{(2n+2)!}$

Then $I_{n} = \sum_{i =0}^{n-1} \frac{2^{2i+1}i!(i+1)!}{(2i+2)!} + 1$

Not sure how useful it is, but I'm pretty sure it's correct

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