$\int \cos ^3 x e^{\log (\sin x)} d x$ is equal to

$-\frac{\sin ^4 x}{4}+c$ $-\frac{\cos^4 x}{4}+c$ $\frac{e^{\sin x}}{4}+c$ none of these

$-\frac{\cos^4 x}{4}+c$

$e^{\log \sin x}=\sin x$ ∴   $\int \cos ^3 x \sin x d x$  Put cos x = t $\Rightarrow-\int t^3 d t=-\frac{t^4}{4}+c=-\frac{\cos ^4 x}{4}+c$ Hence (2) is the correct answer.