Answer
$-\frac{4}{3}\cos^3(\frac{1}{2}x)+C$
Work Step by Step
$\int\sin x\cos(\frac{1}{2}x)dx$
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 $\sin u\ du=-dv$.
$=4\int v^2*-1 dv$
$=-4*\frac{v^3}{3}+C$
$=-\frac{4}{3}\cos^3 u+C$
$=\boxed{-\frac{4}{3}\cos^3(\frac{1}{2}x)+C}$