$\int\dfrac{1+\cos^5 x}{1+\sin x}dx=\int\dfrac{1}{1+\sin x}dx+\int\dfrac{\cos^4 x d(\sin x)}{1+\sin x}=I_1+I_2$
$I_1 = \dfrac{1}{(\sin \dfrac{x}{2}+\cos \dfrac{x}{2})^2}dx=\dfrac{1}{2}\int \dfrac{1}{\sin^2 (\dfrac{x}{2}+\dfrac{\pi}{4})}dx=-\cot (\dfrac{x}{2}+\dfrac{\pi}{4})+C$
$I_2=\int \dfrac{(1-\sin^2 x)^2 d(\sin x)}{1+\sin x}=\int \dfrac{(1-t^2)^2}{1+t}dt=\int (1-t)^2 (1+t) dt$ dễ rồi nhé