Question

Evaluate:$\int (\sqrt{cotx}+\sqrt{tanx})dx$

Answer 1

$\text{Let}$ $I=\int (\sqrt{cotx}+\sqrt{tanx})dx,$

$=\int (t+\frac{1}{t})\frac{2t}{1+t^4}dtPut[\sqrt{tanx}=t,tanx=t^2$
$se{c}^{2}xdx=2tdt$
$\Rightarrow (1+t^4)dx=2tdt$
$\Rightarrow dx=\frac{2t}{1+t^4}dt]$
$=2\int \frac{t^2+1}{t^4+1}dt$

$\text{[Divide numerator and denominator by}{t}^2\text{]}$

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

[$\text{Put}$ $t-\frac{1}{t}=u,t^2+\frac{1}{t^2}-2=u^2$ $(1+\frac{1}{t^2})dt=du]$

$=2\int \frac{du}{2+u^2}=\frac{2}{\sqrt{2}}ta{n}^{-1}\frac{u}{\sqrt{2}}$

$=\frac{2}{\sqrt{2}}ta{n}^{-1}\left(\frac{t-\frac{1}{t}}{\sqrt{2}}\right)=\frac{2}{\sqrt{2}}ta{n}^{-1}\left(\frac{tanx-1}{\sqrt{2}\sqrt{tanx}}\right)+c$