1 Answers
1I=\int {\frac{1}{(2sinx+secx)^4}dx} \\\\=\int {\frac{sec^4x}{(2tanx+sec^2x)^4}dx} \\\\=\int {\frac{(1+tan^2x)sec^2x}{(2tanx+tan^2x+1)^4}dx}=\int {\frac{(1+tan^2x)sec^2x}{(1+tanx)^8}dx} \\ \\ \\Now\: take\: 1+tanx=t \\ \\I=\int{\frac{t^2-2t+2}{t^8}dt}
Integral can be evaluated easily now.
