Question

$\int x{\sin }^{3}xdx$.

Answer 1

$\int {\sin }^{3}xdx$

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

$-\cos x-\int {\cos }^{2}x\sin xdx$

$-\cos x+\int u^{2}du$

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

Integrating by parts, we get

$x(-\cos x+\frac{{\cos }^{3}x}{3})-\int (-\cos x+\frac{{\cos }^{3}x}{3})dx$

$-x\cos x+x\frac{{\cos }^{3}x}{3}+\sin x-\int {\cos }^{3}xdx$

$-x\cos x+x\frac{{\cos }^{3}x}{3}+\sin x-\int \cos x(1-{\sin }^{2}x)dx$

$-x\cos x+x\frac{{\cos }^{3}x}{3}+\sin x-\int \cos xdx-\int \cos x{\sin }^{2}xdx$

$-x\cos x+x\frac{{\cos }^{3}x}{3}-\int u^{2}du$

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

$-x\cos x+x\frac{{\cos }^{3}x}{3}-\frac{{\sin }^{3}x}{3}$