$\int\frac{n\cos^{n-1}}{(1+\sin x)^n}dx$ equals:

$\frac{\cos^nx}{(1+\sin x)^n}$ $\frac{\sin^nx}{(1+\sin x)^n}$ $\frac{-\cos^nx}{(1+\sin x)^n}$ $(\sec x+\tan x)^{-n}+C$

$\frac{-\cos^nx}{(1+\sin x)^n}$

$I=\int\frac{n\cos^{n-1}}{(1+\sin x)^n}dx=\int\frac{n\sec x\,dx}{(\sec x+tan x)^n}=\int{n(\sec x)(\sec x+\tan xdx)}{(\sec x+tan x)^{n+1}}$ Put $(\sec x+tan x)=t$ and $\sec x(\sec x+tan x)dx=dt$ $⇒I=\int\frac{n\,dt}{t^{n+1}}=\frac{-1}{t^n}=\frac{-1}{(\sec x+tan x)^n}=\frac{-\cos^nx}{(1+\sin x)^n}$