Integral exercise asked by Katrina Linda intehral of squareroot tangent

Exercise:

Compute, $\int_{0}^{\frac{\pi }{2}}{\sqrt{\tan x}\,dx}$
Solution: we have $I=\int_{0}^{\frac{\pi }{2}}{\sqrt{\tan x}\,dx}\Leftrightarrow I=\int_{0}^{\frac{\pi }{2}}{\sqrt{\tan \left( \frac{\pi }{2}-x \right)}\,dx}=\int_{0}^{\frac{\pi }{2}}{\sqrt{\cot x}\,dx}$

So $2I=\int_{0}^{\frac{\pi }{2}}{\left( \sqrt{\tan x}+\sqrt{\cot x} \right)dx}=\int_{0}^{\frac{\pi }{2}}{\left( \frac{\sqrt{\sin x}}{\sqrt{\cos x}}+\frac{\sqrt{\cos x}}{\sqrt{\sin x}} \right)dx}\Leftrightarrow 2I=\int_{0}^{\frac{\pi }{2}}{\frac{\sin x+\cos x}{\sqrt{\sin x\cos x}}dx}$

But $\sqrt{\sin x\cos x}=\sqrt{\frac{1}{2}2\sin x\cos x}=\frac{1}{\sqrt{2}}\sqrt{\sin 2x}=\frac{\sqrt{\sin 2x}}{\sqrt{2}}$

So $2I=\int_{0}^{\frac{\pi }{2}}{\frac{\sin x+\cos x}{\frac{\sqrt{\sin x}}{\sqrt{2}}}dx}=\sqrt{2}\int_{0}^{\frac{\pi }{2}}{\frac{\sin x+\cos x}{\sqrt{\sin 2x}}dx}$

Observe that , $\sin 2x=\sin 2x+1-1=2\sin x\cos x-{{\cos }^{2}}x-{{\sin }^{2}}x+1$

                            $=-\left( {{\cos }^{2}}x+{{\sin }^{2}}x-2\sin x\cos x \right)+1=1-{{\left( \cos x-\sin x \right)}^{2}}$

So $I=\frac{\sqrt{2}}{2}\int_{0}^{\frac{\pi }{2}}{\frac{\sin x+\cos x}{\sqrt{1-{{\left( \cos x-\sin x \right)}^{2}}}}dx}=-\frac{\sqrt{2}}{2}\int_{0}^{\frac{\pi }{2}}{\frac{d\left( \cos x-\sin x \right)}{\sqrt{1-{{\left( \cos x-\sin x \right)}^{2}}}}}$

Thus $I=-\frac{\sqrt{2}}{2}\left( \arcsin \left( \cos x-\sin x \right) \right)_{0}^{\frac{\pi }{2}}=-\frac{\sqrt{2}}{2}\arcsin \left( -1 \right)-\arcsin \left( 0 \right)=\frac{\sqrt{2}}{2}\times \frac{\pi }{2}=\frac{\pi }{\sqrt{2}}$