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Show: \begin{align*} \int\limits_0^1\sqrt{\frac{1-x^2}{1+x^2}}\,\mathrm d x &=\frac{\sqrt \pi}{4}\left(\frac{\Gamma \left(\frac14\right)}{\Gamma\left(\frac34\right)}-4\frac{\Gamma\left(\frac34\right)}{\Gamma\left(\frac14\right)}\right) \end{align*}


First attempt: \begin{align*} \int\limits_0^1\sqrt{\frac{1-x^2}{1+x^2}}\,\mathrm d x &=\int\limits_0^1{\left(1-x^2\right)}^{1/2}{\left(1+x^2\right)}^{-1/2}\,\mathrm d x\\ \qquad\text{let } x=u^{1/2}\\ \,\mathrm d x =\frac 1 2 u ^{-1/2}\,\mathrm d u\\ &=\frac 1 2 \int\limits_0^1{\left(1-u\right)}^{1/2}{\left(1+u\right)}^{-1/2}u^{-1/2}\,\mathrm d u\\\qquad \text{let }1-u= y\\ u= 1-y,\\ \,\mathrm du=-\,\mathrm dy\\ &=-\frac 1 2 \int_0^1y^{1/2}\left(2-y\right)^{-1/2}(1-y)^{-1/2}\,\mathrm d y\\ \\&=\quad\color{orange}{????} \\ \end{align*}


Second attempt: \begin{align*} \\&\int\limits_0^1\sqrt{\frac{1-x^2}{1+x^2}}\,\mathrm dx\\ x^2=\cos(2\theta)\\ \mathrm dx=\frac{-\sin(2\theta)}x\,\mathrm d\theta\\ x=0\rightarrow\theta=\pi\\ x=1\rightarrow\theta=0\\ &=\int\limits_\pi^0\sqrt{\frac{1-\cos(2\theta)}{1+\cos(2\theta)}}\cdot\frac{-\sin(2\theta)}{\sqrt{\cos(2\theta)}}\,\mathrm d\theta\\ 2\sin^2(x)=1-\cos(2x)\\ 2\cos^2(x)=1+\cos(2x)\\ \sin(2x)=2\sin(x)\cos(x)\\ &=\int\limits_\pi^0\sqrt{\frac{2\sin^2\theta }{2\cos^2\theta}}\cdot\frac{-2\sin(\theta)\cos(\theta)}{\sqrt{\cos(2\theta)}}\,\mathrm d\theta\\ &=2\int\limits^\pi_0\tan(\theta)\cdot\frac{\sin(\theta)\cos(\theta)}{\sqrt{\cos(2\theta)}}\,\mathrm d\theta\\ &=2\int\limits^\pi_0\frac{\sin^2(\theta)}{\sqrt{\cos(2\theta)}}\,\mathrm d\theta\\ \text{let } \theta=2\phi\\ \mathrm d\theta =2\mathrm d\phi\\ \theta=0\rightarrow\phi=0\\\theta=\pi\rightarrow\phi=\pi/2\\ &=2\int\limits_0^{\pi/2}\frac{\sin^2(2\phi)}{\sqrt{\cos(4\phi)}}2\,\mathrm d\phi\\ &=4\int\limits_0^{\pi/2}\frac{\sin^2(2\phi)}{\sqrt{\cos(4\phi)}}\,\mathrm d\phi\\ \\& =\quad\color{green}{????}\\ \end{align*}


@doraemonpaul's Method: \begin{align*} \int\limits_0^1\sqrt{\frac{1-x^2}{1+x^2}}\,\mathrm dx &=\int\limits_0^1\sqrt{\frac{1-x^2}{1+x^2}}\times\sqrt{\frac{1-x^2}{1-x^2}}\,\mathrm dx\\ &=\int\limits_0^1\frac{1-x^2}{\sqrt{1-x^4}}\,\mathrm dx\\ &=\int\limits_0^1(1-x^4)^{-1/2}\,\mathrm dx-\int\limits_0^1x^2(1-x^4)^{-1/2}\,\mathrm dx\\ \text{let }x=u^{1/4}\\ \mathrm dx=\frac14 u^{-3/4}\,\mathrm du\\ &=\int\limits_0^1(1-u)^{-1/2}\cdot\frac14 u^{-3/4}\,\mathrm du-\int\limits_0^1u^{1/2}(1-u)^{-1/2}\cdot\frac14 u^{-3/4}\,\mathrm du\\ &=\frac14\int\limits_0^1u^{-3/4}(1-u)^{-1/2}\cdot \,\mathrm du-\frac14\int\limits_0^1u^{-1/4}(1-u)^{-1/2}\,\mathrm du\\ \small{\text B(m,n)=\int\limits_0^1t^{m-1}(1-t)^{n-1} \,\mathrm d t}\\ \small{m_1-1=-3/4\quad n_1-1}&\small{=-1/2\quad m_2-1=-1/4\quad n_2-1=-1/2}\\ \small{m_1=1/4\quad \quad n_1=1/2\quad}&\small{\quad m_2=3/4\quad\quad n_2=1/2}\\ \\ &=\frac14\text B\left(\tfrac 14,\tfrac12\right)-\frac14\text B\left(\tfrac34,\tfrac12\right)\\ \small{\text B(m,n)=\frac{\Gamma(m)\Gamma(n)}{\Gamma(m+n)}}\\ &=\frac 14\left(\frac{\Gamma(\frac 14)\Gamma(\frac 12)}{\Gamma(\frac 34)}-\frac{\Gamma(\frac34)\Gamma(\frac 12)}{\Gamma(\frac 54)}\right)\\ &=\frac {\sqrt\pi}4\left(\frac{\Gamma(\frac 14)}{\Gamma(\frac 34)}-\frac{\Gamma(\frac34)}{\Gamma(\frac 54)}\right)\\ &=\frac {\sqrt\pi}4\left(\frac{\Gamma(\tfrac 14)}{\Gamma(\tfrac 34)}-\frac{\Gamma(\tfrac34)}{\tfrac 14\Gamma(\tfrac 14)}\right)\\ &=\frac{\sqrt\pi}4\left(\frac{\Gamma(\tfrac 14)}{\Gamma(\tfrac 34)}-4\frac{\Gamma(\tfrac34)}{\Gamma(\tfrac 14)}\right)\\ \end{align*}

NB: this question has been edited post comments ,

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You generally have to put more effort asking questions here. At least try to show that you tried and failed, what you tried, what you know, etc. Also, writing a little bit of english is not a bad thing... This seems to be homework and you seemed not to be bothered. –  chango Nov 3 '12 at 10:55
    
im new to this space –  Elements in Space Nov 3 '12 at 10:59
3  
Ok, I am just letting you know. By the way, there seems to be an error in the second equality. You need to add a $u^{-1/2}$ multiplying the whole integrand when you do that substitution. –  chango Nov 3 '12 at 11:01
    
wolfram alpha's result –  doraemonpaul Nov 4 '12 at 3:44

1 Answer 1

up vote 3 down vote accepted

I think you treat this question too complicated.

First it should note that integral of the form $\int_0^1(1-x^2)^a(1+x^2)^b~dx$ is impossible to have easy-look substitution to transform it to a integral of the form $\int_0^1x^p(1-x)^q~dx$ (i.e. beta function).

Second it should note that trigonometric substitution also often bring this type of question too complicated.

But I admit that this question has a level of lucky or tricky.

$\int_0^1\sqrt{\dfrac{1-x^2}{1+x^2}}~dx$

$=\int_0^1\dfrac{1-x^2}{\sqrt{1-x^2}\sqrt{1+x^2}}~dx$

$=\int_0^1\dfrac{1-x^2}{\sqrt{1-x^4}}~dx$

$=\int_0^1(1-x^4)^{-\frac{1}{2}}~dx-\int_0^1x^2(1-x^4)^{-\frac{1}{2}}~dx$

$=\int_0^1(1-x)^{-\frac{1}{2}}~d(x^{\frac{1}{4}})-\int_0^1x^{\frac{1}{2}}(1-x)^{-\frac{1}{2}}~d(x^{\frac{1}{4}})$

$=\dfrac{1}{4}\int_0^1x^{-\frac{3}{4}}(1-x)^{-\frac{1}{2}}~dx-\dfrac{1}{4}\int_0^1x^{-\frac{1}{4}}(1-x)^{-\frac{1}{2}}~dx$

$=\dfrac{B\left(\dfrac{1}{4},\dfrac{1}{2}\right)}{4}-\dfrac{B\left(\dfrac{3}{4},\dfrac{1}{2}\right)}{4}$

$=\dfrac{\Gamma\left(\dfrac{1}{4}\right)\Gamma\left(\dfrac{1}{2}\right)}{4\Gamma\left(\dfrac{3}{4}\right)}-\dfrac{\Gamma\left(\dfrac{3}{4}\right)\Gamma\left(\dfrac{1}{2}\right)}{4\Gamma\left(\dfrac{5}{4}\right)}$

$=\dfrac{\sqrt{\pi}~\Gamma\left(\dfrac{1}{4}\right)}{4\Gamma\left(\dfrac{3}{4}\right)}-\dfrac{\sqrt{\pi}~\Gamma\left(\dfrac{3}{4}\right)}{\Gamma\left(\dfrac{1}{4}\right)}$

$=\dfrac{\sqrt{\pi}~\Gamma\left(\dfrac{1}{4}\right)}{4\times\dfrac{\sqrt{2}\pi}{\Gamma\left(\dfrac{1}{4}\right)}}-\dfrac{\sqrt{\pi}~\Gamma\left(\dfrac{3}{4}\right)}{\dfrac{\sqrt{2}\pi}{\Gamma\left(\dfrac{3}{4}\right)}}$

$=\dfrac{1}{4\sqrt{2\pi}}\Gamma^2\left(\dfrac{1}{4}\right)-\dfrac{1}{\sqrt{2\pi}}\Gamma^2\left(\dfrac{3}{4}\right)$

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