$∫sin^3x\, dx$ is equal to :

$\frac{cos^3x}{3}-cos\, x+C$

The correct answer is Option (2) → $\frac{\cos^3x}{3}-\cos x+C$

$∫\sin^3x dx=\int\sin x.\sin^2xdx$

$=\int\sin x(1-\cos^2x)dx$

$=\sin xdx-\int\sin x\cos^2x dx$

$=-\cos x-\int\sin x\cos^2x dx$

let $u=\cos x$

$⇒\frac{du}{dx}=-\sin x⇒\frac{du}{-\sin x}=dx$

$∴\int\sin^3xdx=-\cos x+\int u^2du$

$=-\cos x+\frac{u^3}{3}+C$

$=-\cos x+\frac{\cos^3x}{3}+C$