$\int \sin x d(\cos x)$ is equal to

$\frac{1}{2}\left(\frac{1}{2} \sin 2 x-x\right)+C$

We have,

$d(\cos x)=\frac{d}{d x}(\cos x) \cdot d x=-\sin x d x$

∴  $\int \sin x d(\cos x)=-\int \sin ^2 x d x=-\frac{1}{2} \int(1-\cos 2 x) d x$

⇒  $\int \sin x d(\cos x)=-\frac{1}{2}\left(x-\frac{\sin 2 x}{2}\right)+C$

$=\frac{1}{2}\left(\frac{1}{2} \sin 2 x-x\right)+C$