Show that

Question:

$\cos ^{3} x e^{\log \sin x}$

Solution:

$\cos ^{3} x e^{\log \sin x}=\cos ^{3} x \times \sin x$

Let $\cos x=t \Rightarrow-\sin x d x=d t$

$\Rightarrow \int \cos ^{3} x e^{\log \sin x} d x=\int \cos ^{3} x \sin x d x$

$=-\int t \cdot d t$

$=-\frac{t^{4}}{4}+C$

$=-\frac{\cos ^{4} x}{4}+C$

Leave a comment