Question

Solve it in a shortest possible way.​

Answer 1

Answer:

$\large\boxed{\sf{I= { \tan }^{ - 1} ( \sin x) + \frac{1}{2} ln(1 + { \sin}^{2} x) - ln(1 - \sin x)+C }}$

Step-by-step explanation:

Given

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

To solve it further, let's assume that

$\sin(x) = t \\ \\ = > \cos(x) dx = dt$

Therefore , we get,

$\displaystyle \int \frac{2}{(1 - t)(1 + {t}^{2} )} dt$

Now, it will be solved by integrating partial fraction.

• $\bold\red{Refer\: to \:the\: attachment .}$

Thus, we get the answer as :

$I= \purple{{ \tan }^{ - 1} ( \sin x) + \frac{1}{2} ln(1 + { \sin}^{2} x) - ln(1 - \sin x)+C }$