Đặt $\ln (\sin x ) = u \Rightarrow \dfrac{\cos x}{\sin x} dx=du$ và $\sin x dx =dv \Rightarrow -\cos x =v$
$I=-\cos x.\ln (\sin x) +\int \cos x \dfrac{\cos x}{\sin x}dx=-\cos x.\ln (\sin x) +\int \dfrac{1-\sin^2 x}{\sin x}dx$
$=-\cos x.\ln (\sin x) +\int \dfrac{1}{\sin x}dx -\int \sin x dx =-\cos x.\ln (\sin x) +\cos x +\int \dfrac{\sin x}{\sin^2 x}dx$
Tính $I_1=\int \dfrac{\sin x}{\sin^2 x}dx=-\int \dfrac{d(\cos x)}{1-\cos^2 x}=-\int\dfrac{1}{1-t^2}dt =-\dfrac{1}{2}\int (\dfrac{1}{1-t}+\dfrac{1}{1+t})dt$
$=-\dfrac{1}{2}\ln |\dfrac{t+1}{1-t}| +C$
Lắp lại xong nhé