Question

Evaluate:$\underset{0}{\overset{\pi /2}{\int }}\left(\sqrt{\tan x}+\sqrt{\cot x}\right)dx$

Answer 1

${}^{\pi /2}{\int }_0\sqrt{\tan x}+\sqrt{\cot x}dx$

$=\int \sqrt{\tan x}dx+\int \sqrt{\cot x}dx$

$=\int {\sec }^{2}x\frac{\sqrt{\tan x}}{1+{\tan }^{2}x}dx+\int cose{c}^{2}x\frac{\sqrt{\cot x}}{1+{\cot }^{2}x}dx$

Solving separately,

$u=\tan x$ $v=\cot x$

$=\int \frac{\sqrt{u}}{u^2+1}du+\int \frac{\sqrt{v}}{v^2+1}dv$

$v=\sqrt{u}$

$\to du=2\sqrt{u}dv$

$=2\int \frac{v^2}{v^u+1}dv+2\int \frac{v^2}{v^u+1}dv$

Solving the integration & applying the limits we gets

${}^{\pi /2}{\int }_0\sqrt{\tan x}+\sqrt{\cot x}dx=\sqrt{2}\pi$.