Question

$\int \frac{1}{1+\sin x}dx=$

Answer 1

Now,

$\int \frac{1}{1+\sin x}dx$

$=\int \frac{1-\sin x}{1-{\sin }^{2}x}dx$

$=\int \frac{1-\sin x}{{\cos }^{2}x}dx$

$=\int ({\sec }^{2}x-\tan x.\sec x)dx$

$=\tan x-\sec x+c$. [ Where $c$ is the integrating constant]