Question

integration of tan inverse(cosecx - cotx) dx​

Answer 1

Step-by-step explanation:

some rule to solve this question :

sinx/2 / cosx/2 = tanx/2

tan inverse ( tanx/2 ) = x/2

Answer 2

Explanation:

$I = \int \tan^-^1 (\csc x +\cot x)dx$

$I = \int \tan^-^1 (\csc x +\cot x)dx\\\\taking\\\\\tan^-^1 (\csc x +\cot x)dx\\\\\tan^-^1\left ( \frac{1}{\sin x}+\frac{\cos x}{\sin x} \right )\\\\\text{taking LCM }\\\\\tan^-^1\left ( \frac{1+\cos x}{\sin x} \right )$

$\tan^-^1 \left ( \frac{2\cos^2\frac{x}{2}}{2\sin^2\frac{x}{2} 2\cos^2\frac{x}{2}} \right )$

we know that $\cot x = \frac{\cos x}{\sin x}$

thus

$\tan^-^1 \left \left ( \cot \frac{x}{2} \right )= \tan^-^1(\tan (\frac{\pi}{2}-\frac{x}{2}))\\\\=\frac{\pi}{2}-\frac{x}{2}$

$then \\\\I = \int (\frac{\pi}{2}-\frac{x}{2})dx\\\\\frac{\pi}{2}x -\frac{x^2}{4}+c$