$\int\frac{n\cos^{n-1}}{(1+\sin x)^n}dx$ equals:
Answer & explanation
Correct answer: option 3
$I=\int\frac{n\cos^{n-1}}{(1+\sin x)^n}dx=\int\frac{n\sec x\,dx}{(\sec x+tan x)^n}=\int{n(\sec x)(\sec x+\tan xdx)}{(\sec x+tan x)^{n+1}}$
Put $(\sec x+tan x)=t$ and $\sec x(\sec x+tan x)dx=dt$
$⇒I=\int\frac{n\,dt}{t^{n+1}}=\frac{-1}{t^n}=\frac{-1}{(\sec x+tan x)^n}=\frac{-\cos^nx}{(1+\sin x)^n}$