Question

please solve this question​

Answer 1

Step-by-step explanation:

We have,

$\int \tan^{ - 1} ( \frac{1 - \tan(x) }{1 + \tan(x) } ) dx$

$= \int \tan^{ - 1} ( \frac{ \tan( \frac{\pi}{4} ) - \tan(x) }{ \tan( \frac{\pi}{4} ) + \tan(x) } ) dx$

$= \int \tan^{ - 1} ( \tan( \frac{\pi}{4} - x ) ) dx$

$= \int( \frac{\pi}{4} - x)dx$

$= \frac{\pi}{4}x - \frac{ {x}^{2} }{2} + c$