Integrals — Class 12 Maths Solution

Let $I = \int {{{\cos }^3}x{e^{\log \sin x}}dx} = \int {{{\cos }^3}x\sin xdx}$

Put $\cos x = t$ $\Rightarrow$ $- \sin xdx = dt$

$\therefore$ $I = - \int {{t^3}} dt = - \cfrac{{{t^4}}}{4} + C = - \cfrac{1}{4}{\cos ^4}x + C$