A calculus problem by Nishanth Subramanian

This is not a detailed solution but I will be highlighting the main steps involved

Put $x^2 =\tan{\theta}$ The integral reduces to

$\displaystyle \int_0^{\frac{\pi}{4}}\mathrm{\frac{\sqrt{\cot{\theta}}~d\theta}{2}}$

This can be written as

$\displaystyle \int_0^{\frac{\pi}{4}}\mathrm{\frac{ (\sqrt{\cot{\theta}} + \sqrt{\tan{\theta}} ) + (\sqrt{\cot{\theta}} - \sqrt{\tan{\theta}}) d\theta}{4}}$

$I_1 = \displaystyle \int_0^{\large\frac{\pi}{4}}\mathrm{ \frac{\sqrt{\cot{\theta}} + \sqrt{\tan{\theta}}~d\theta}{4}} = \displaystyle \int_0^{\large\frac{\pi}{4} }\mathrm{\frac{\sin{\theta} +\cos{\theta}~d\theta}{4\sqrt{\sin{\theta}\cos{\theta}}}}$

Substitute $\sin{\theta} - \cos{\theta} = t$ The numerator is $dt$ .The denominator can be easily expressed in terms of $t$

$\displaystyle \sin{\theta}\cos{\theta} = \frac{ 1- t^2}{2}$

Similarly for $I_2$ = $\displaystyle \int_0^{\frac{\pi}{4}}\mathrm{\frac{\sqrt{\cot{\theta}} - \sqrt{\tan{\theta}} ~d\theta}{4}} = \int_0^{\frac{\pi}{4}}\mathrm{\frac{\cos{\theta} - \sin{\theta}~d\theta}{4\sqrt{\sin{\theta}\cos{\theta}}}}$ , substitute $\sin{\theta} + \cos{\theta} = m$

The required integral is $\color{#20A900}{ I_1 + I_2 }$

After evaluating the integrals,this comes out to be

$\displaystyle \color{#3D99F6}{\frac{ \ln{(\sqrt{2} + 1)} +\frac{\pi}{2} }{ 2\sqrt{2}} = \boxed{0.86697} }$