$I=\int \dfrac{\cos x \sin x}{(1+\cos x)\sin^2 x}dx =-\int \dfrac{\cos x }{(1+\cos x)(1-\cos^2 x)}d(\cos x)$
$=-\int \limits_{\frac{\sqrt 3}{2}}^{0} \dfrac{t}{(1+t)^2 (1-t)}dt =-\int \bigg (\dfrac{1}{4(t+1)}-\dfrac{1}{2(t+1)^2}-\dfrac{1}{4(t-1)}\bigg )dt$
$=\bigg (\dfrac{1}{4} \bigg[\ln |t-1|-\ln |t+1| \bigg ] +\dfrac{1}{2(t+1)}\bigg ) \bigg |_{\frac{\sqrt 3}{2}}^{0}=...$