Ta có : \(I=\int \frac{\cos x}{\left(\sin x+\cos x\right)^{3} } {\rm d}x =\int \frac{{\rm d}x}{\cos ^{2} x.\left(1+\tan x\right)^{3} }\) .
Đặt \( t=1+\tan x\Rightarrow {\rm dt}=\frac{1}{\cos ^{2} x} {\rm d}x.\)
Suy ra: \(I=\int \frac{{\rm dt}}{t^{3} } =-\frac{1}{2} \cdot \frac{1}{t^{2} } +C.\)
Vậy \(I=-\frac{1}{2} \cdot \frac{1}{\left(1+\tan x\right)^{2} } +C.\)