How to evaluate $\int\frac{dx}{(2\sin x+\sec x)^4}$?

Question

I tried a lot but I am not able to get a start.

Can anyone give me the start of this question $$ \int\frac{dx}{(2\sin x+\sec x)^4} \ ? $$

Answer 1

One may start with the change of variable $$ t=\tan x,\quad \cos (2x)=\frac{1-t^2}{1+t^2},\quad \sin (2x)=\frac{2t}{1+t^2}, \quad dx=\frac1{1+t^2}\:dt, $$ giving $$ \begin{align} \int\frac{dx}{(2\sin x+\sec x)^4}&=\int\frac{\cos^4 x\:dx}{(2\sin x\cos x+1)^4} \\\\&=\frac14\int\frac{(1+\cos (2x))^2}{(\sin (2x)+1)^4}\:dx \\\\&=2\int\frac{1+t^2}{(t+1)^8}\:dt. \end{align} $$

Answer 2

$$\int\frac{dx}{(2\sin x+\sec x)^4} $$ $$\int\frac{\sec^4 x}{(2\tan x + 1 + \tan^2 x)^4} $$ Put $\tan x +1 =t$ $$\int\frac{(t-1)^2 +1 }{t^8} $$