Question

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

Answer 1

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

$\Rightarrow \int \frac{1+tanx}{\sqrt{tanx}}dx$

Put $tanx=z^2\Rightarrow se{c}^{2}xdx=2zdz$

$dx=\frac{2zdz}{1+z^4}$

$I=\int \frac{1+z^2}{z}.\frac{2zdz}{1+z^4}=2\int \frac{z^2+1}{z^4+1}dz$

$=2\int \frac{1+\frac{1}{z^2}}{z^2+\frac{1}{z^2}}dz$

$=2\int \frac{1+\frac{1}{z^2}}{{(z-\frac{1}{z})}^2+2}dz$

Put $z-\frac{1}{z}=u$

$\Rightarrow (1+\frac{1}{z^2})dz=du$

$I=2\int \frac{du}{u^2+(\sqrt{2}{)}^2}$

$I=2.\frac{1}{\sqrt{2}}ta{n}^{-1}\left(\frac{u}{\sqrt{2}}\right)+C=\sqrt{2}ta{n}^{-1}\left(\frac{z-\frac{1}{z}}{\sqrt{2}}\right)+C$

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

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