Question

Evaluate $\int \frac{1}{1+\sin 2x}dx$

Answer 1

$\int \frac{1}{1+\sin 2x}dx$

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

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

$=\int \frac{{\sec }^{2}x}{(\tan x+1{)}^2}dx$......(1)

Put $1+\tan x=z\Rightarrow {\sec }^{2}xdx=dz$. Using these from (1) we get,

$=\int \frac{dz}{z^2}dx$

$=-\frac{1}{z}+c$ [Where $c$ is integrating constant]

$=-\frac{1}{1+\tan x}+c$