Question

integrate 1by 1+ tanx dx​

Answer 1

Step-by-step explanation:

We have,

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

$= \int \frac{ \cos(x)dx }{ \sin(x) + \cos(x) }$

$= \int \frac{( \frac{1}{2}( \cos(x) + \sin(x) ) + \frac{1}{2} ( \cos(x) - \sin(x))) dx}{ \sin(x) + \cos(x) }$

$= \frac{1}{2} \int \: dx + \frac{1}{2} \int \frac{ \cos(x) - \sin(x) }{ \cos(x) + \sin(x) } dx$

Let, sin(x) + cos(x) = t => {cos(x) - sin(x)}dx = dt

$= \frac{x}{2} + \frac{1}{2} \int \frac{dt}{t}$

$= \frac{x}{2} + \frac{1}{2} \: ln(t) + c$

$= \frac{x}{2} + ln( \sqrt{ \sin(x) + \cos(x) } ) + c$