Question

Solve : $\int \sqrt{\tan x}.dx$

Answer 1

put $\tan x=t^2$

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

$(1+{\tan }^{2}x)dx=2tdt$

$dx=\frac{2tdt}{1+t^4}$

$\int \sqrt{\tan x}.dx$

$\int t\times \frac{2t}{1+t^4}dt$

$2\int \frac{t^2}{1+t^4}dt$

$\int \frac{\left[\right(t^2+1)+(t^2-1\left)\right]}{(1+t^4)}dt$

$\Rightarrow \int (t^2+1)/(1+t^4)dt+\int (t^2-1)/(1+t^4)dt$

$\Rightarrow \int (1+1/t^2)/(t^2+1/t^2)dt+\int (1-1/t^2)/(t^2+1/t^2)dt$

$\Rightarrow \int (1+1/t^2)dt/\left[\right(t-1/t{)}^2+2]+\int (1-1/t^2\left)dt/\right[(t+1/t{)}^2-2]$

Let t−1/t=u for the first integral $\Rightarrow (1+1/t^2)dt=du$

and t+1/t=v for the 2nd integral $\Rightarrow (1-1/t^2)dt=dv$

Integral

$=\int du/(u^2+2)+\int dv/(v^2-2)$

$=\left(1/\surd 2\right)tan-1\left(u/\surd 2\right)+\left(1/2\surd 2\right)log(v-\surd 2)/(v+\surd 2)l+c$

$=\left(1/\surd 2\right)tan-1\left[\right(t^2-1\left)/t\surd 2\right]+\left(1/2\surd 2\right)log(t^2+1-t\surd 2)/t^2+1+t\surd 2)+c$

$\left[\frac{1}{\sqrt{2}}{\tan }^{-1}(\frac{\tan x-1}{\sqrt{2}\tan x}+\frac{1}{2\sqrt{2}}\log \left|\frac{\tan x+1-\sqrt{2}\sqrt{\tan x}}{\tan x-1+\sqrt{2}\sqrt{\tan x}}\right|)\right]$