Question

integral form of 1-tanx÷1+tanx​

Answer 1

Step-by-step explanation:

We have,

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

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

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

$= \frac{ ln | \sec( \frac{\pi}{4} - x ) | }{ - 1} + c$

$= - ln |\sec( \frac{\pi}{4} - x ) |+c$

$= ln | \cos( \frac{\pi}{4} - x ) | +c$