Question

Integrate the following
$int \: \: tan {}^{ - 1} \sqrt{ \frac{1 + cos \: x}{1 -cos \: x} } dx$
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Answer 1

Answer:

Step-by-step explanation:

We have,

$\displaystyle\int\tan^{-1}\left(\sqrt{\dfrac{1+\cos(x)}{1-\cos(x)}}\right)\,dx$

$\displaystyle=\int\tan^{-1}\left(\sqrt{\dfrac{2\cos^{2}\left(\dfrac{x}{2}\right)}{2\sin^{2}\left(\dfrac{x}{2}\right)}}\right)\,dx$

$\displaystyle=\int\tan^{-1}\left(\sqrt{\dfrac{\cos^{2}\left(\dfrac{x}{2}\right)}{\sin^{2}\left(\dfrac{x}{2}\right)}}\right)\,dx$

$\displaystyle=\int\tan^{-1}\left(\sqrt{\cot^{2}\left(\dfrac{x}{2}\right)}\right)\,dx$

$\displaystyle=\int\tan^{-1}\left(\cot\left(\dfrac{x}{2}\right)\right)\,dx$

$\displaystyle=\int\tan^{-1}\left(\tan\left(\dfrac{\pi}{2}-\dfrac{x}{2}\right)\right)\,dx$

$\displaystyle=\int\left(\dfrac{\pi}{2}-\dfrac{x}{2}\right)\,dx$

$=\dfrac{\pi x}{2}-\dfrac{{x}^{2}}{4}+C$