Evaluate the integral : d

Question

Evaluate the integral : $\int \sqrt{\cot \theta} d \theta$

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Answer

$I=\int \sqrt{\cot \theta} \ d \ \theta$
Let $\cot \theta=x^2$
$\Rightarrow-\operatorname{cosec} 2 \theta \ d \ \theta=2 x \ dx$
$\Rightarrow d\ \theta=\frac{-2 x}{\operatorname{cosec}^2 \theta} d x$
$=\frac{-2 x}{1+\cot ^2 \theta}\ d x$
$=\frac{-2 x}{1+x^4}\ d x$
$\therefore I=-\int \frac{2 x^2}{1+x^4} d x$
$=-\int \frac{2}{\frac{1}{x^2}+x^2} d x$
Dividing numerator and denominator by $x2$
$=-\int \frac{1+\frac{1}{x^2}+1-\frac{1}{x^2}}{x^2+\frac{1}{x^2}} d x$
$=-\int \frac{\left(1+\frac{1}{x^2}\right)^{d x}}{\left(x-\frac{1}{x}\right)^2+2}-\int \frac{\left(1-\frac{1}{x^2}\right) d x}{\left(x+\frac{1}{x}\right)^2-2}$
Let $ x -\frac{1}{x}= t $
$\Rightarrow\left(1+\frac{1}{x^2}\right) dx = dt$
and $x +\frac{1}{x}= z $
$\Rightarrow\left(1-\frac{1}{x^2}\right) d x= dz$
$\Rightarrow I =-\int \frac{ dt }{t^2+2}-\int \frac{d z}{z^2-2}$
$=-\frac{1}{\sqrt{2}} \tan ^{-1}\left(\frac{t}{\sqrt{2}}\right)-\frac{1}{2 \sqrt{2}} \log \left|\frac{z-\sqrt{2}}{z+\sqrt{2}}\right|+ C$
$=-\frac{1}{\sqrt{2}} \tan ^{-1}\left(\frac{x^2-1}{\sqrt{2} x}\right)-\frac{1}{2 \sqrt{2}} \log \left|\frac{x^2+1-\sqrt{2} x}{x^2+1+\sqrt{2} x}\right|+ C$
$I =-\frac{1}{\sqrt{2}} \tan ^{-1}\left(\frac{\cot \theta-1}{\sqrt{2 \cot \theta}}\right)-\frac{1}{2 \sqrt{2}} \log \left|\frac{\cot \theta+1-\sqrt{2 \cot \theta}}{\cot \theta+1-\sqrt{2 \cot \theta}}\right|+ C $

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