Question

Evaluate: $\int (\sqrt{\tan x}+\sqrt{\cot x})dx$

Answer 1

Let $\sqrt{\tan x}=t$

$\tan x=t^2$

${\sec }^{2}xdx=2tdt$

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

$=\int (t+\frac{1}{t})x\frac{2tdt}{{\sec }^{2}x}$

$=\int (t+\frac{1}{t})\times \frac{2tdt}{1+t^2}(\therefore {\sec }^{2}x=1+{\tan }^{2}x)$

$=\int \frac{t^2+1}{t}\times \frac{2tdt}{t+1}$

$=\int dt=2t+c$

$=2\sqrt{\tan x}+c$