## Calculus: Early Transcendentals 8th Edition

$-\frac{4}{3}\cos^3(\frac{1}{2}x)+C$
$\int\sin x\cos(\frac{1}{2}x)dx$ Let $u=\frac{1}{2}x$. Then $x=2u$. Also, $du=\frac{1}{2}dx$, and $dx=2du$. $=\int\sin(2u)\cos u*2\ du$ $=2\int 2\sin u\cos u*\cos u\ du$ $=4\int\sin u\cos^2 u\ du$ Let $v=\cos u$. Then $dv=-\sin u\ du$, and $\sin u\ du=-dv$. $=4\int v^2*-1 dv$ $=-4*\frac{v^3}{3}+C$ $=-\frac{4}{3}\cos^3 u+C$ $=\boxed{-\frac{4}{3}\cos^3(\frac{1}{2}x)+C}$