## Calculus: Early Transcendentals 8th Edition

$\displaystyle \frac{1}{3}x\cos^{3}x-x\cos x+\frac{2}{3}\sin x+\frac{1}{9}\sin^{3}x+C$
$I=\displaystyle \int x\sin^{3}xdx$ We want to integrate by parts, in which $u=x,\ du=dx$ ... loses x. Since our $dv$ is $\sin^{3}x$ we find $\displaystyle \int\sin^{3}xdx=\int\left(1-\cos^{2}x\right)\sin xdx$ ... From the Strategy for Evaluating $\displaystyle \int\sin^{m}x\cos^{n}xdx$, case (b), ... we use the substitution $\left[\begin{array}{l} u=\cos x,\\ du=-\sin xdx \end{array}\right]$ $=\displaystyle \int\left(1-u^{2}\right)(-du)$ $=\displaystyle \int\left(u^{2}-1\right)du$ $=\displaystyle \frac{1}{3}u^{3}-u+C$ $=\displaystyle \frac{1}{3}\cos^{3}x-\cos x+C$ So, for our integration py parts $\displaystyle \int udv=uv-\int vdu,$ we take $\left[\begin{array}{ll} u=x & dv=\sin^{3}xdx\\ du=x & v=\frac{1}{3}\cos^{3}x-\cos x \end{array}\right]$ $I=\displaystyle \frac{1}{3}x\cos^{3}x-x\cos x-\int\left(\frac{1}{3}\cos^{3}x-\cos x\right)dx$ $=\displaystyle \frac{1}{3}x\cos^{3}x-x\cos x-\frac{1}{3}\int\cos^{3}xdx+\int\cos xdx$ $=\displaystyle \frac{1}{3}x\cos^{3}x-x\cos x-\frac{1}{3}\int\cos^{3}xdx+\sin x$ ... this integral is solved in Example 1. $=\displaystyle \frac{1}{3}x\cos^{3}x-x\cos x-\frac{1}{3}\left(\sin x-\frac{1}{3}\sin^{3}x\right)+\sin x+C$ $=\displaystyle \frac{1}{3}x\cos^{3}x-x\cos x+\frac{2}{3}\sin x+\frac{1}{9}\sin^{3}x+C$