1. ## Integration question

Prove that $\displaystyle \int\limits_{0}^{1}\frac{\sqrt(1-x^2)}{1+x^2}\, dx = \frac{\pi}{2}(\sqrt2-1)$

Hints would be appreciated. Thanks.

2. Originally Posted by anthill
Prove that $\displaystyle \int\limits_{0}^{1}\frac{\sqrt(1-x^2)}{1+x^2}\, dx = \frac{\pi}{2}(\sqrt2-1)$

Hints would be appreciated. Thanks.
substitute 1-x^2=t^2 and express x^2 in denominator using that substitution. You should get a rational function in variable t.

3. Originally Posted by anthill
Prove that $\displaystyle \int\limits_{0}^{1}\frac{\sqrt(1-x^2)}{1+x^2}\, dx = \frac{\pi}{2}(\sqrt2-1)$

Hints would be appreciated. Thanks.
$\displaystyle \displaystyle \int \frac{\sqrt{1-x^2} \cdot \sqrt{1-x^2}}{(1+x^2)\cdot \sqrt{1-x^2}} dx$

$\displaystyle \displaystyle \int \frac{-1 -x^2 }{(1+x^2) \cdot \sqrt{1-x^2}} + \frac{2}{(1+x^2) \cdot \sqrt{1-x^2}} dx$

$\displaystyle \displaystyle \int \frac{-1 -x^2 }{(1+x^2) \cdot \sqrt{1-x^2}} dx +\int \frac{2}{(1+x^2) \cdot \sqrt{1-x^2}} dx$

$\displaystyle \displaystyle \int \frac{-1 }{\sqrt{1-x^2}} dx +\int \frac{2}{(1+x^2) \cdot \sqrt{1-x^2}} dx$

first integration sub
$\displaystyle \sin u = x$
in the second sub
$\displaystyle \displaystyle \tan u = \frac{\sqrt{2} x }{\sqrt{1-x^2}}$

4. Let $\displaystyle x = \sin(y)$

$\displaystyle dx = \cos(y)~dy$

The integral

$\displaystyle = \int_0^{\frac{\pi}{2}} \frac{\cos^2(y)}{ 1+ \sin^2(y) }~dy$

$\displaystyle = \int_0^{\frac{\pi}{2}} \frac{\cos^2(y)}{ \cos^2(y) + 2\sin^2(y) }~dy$

$\displaystyle = \int_0^{\frac{\pi}{2}} \frac{dy}{ 1+ 2\tan^2(y) }$

Consider $\displaystyle 1 = \sec^2(y) - \tan^2(y) - 1/2 + 1/2$ thus we have

$\displaystyle \frac{1}{2} = \sec^2(y) - \frac{1}{2} ( 1+ 2\tan^2(y) )$

The integral

$\displaystyle = 2 \int_0^{\frac{\pi}{2}} \frac{\sec^2(y) }{ 1+ 2\tan^2(y) }~dy - \int_0^{\frac{\pi}{2}} ~dy$

$\displaystyle = \left[ \sqrt{2} \tan^{-1}[\sqrt{2}\tan(y)] - y \right]_0^{\frac{\pi}{2}}$

$\displaystyle = \frac{\pi}{2}( \sqrt{2} - 1 )$