Answer
$\frac{1}{2}\sin(t^2)-\frac{1}{3}\sin^3(t^2)+\frac{1}{10}\sin^5(t^2)$
Work Step by Step
$\int t \cos^5(t^2)dt$
Let $u=t^2$. Then $du=2tdt$, and $tdt=\frac{1}{2}\ du$.
$=\int\cos^5(u)*\frac{1}{2}\ du$
Since the power of cosine is odd, save one factor of $\cos u$ and express the rest in terms of $\sin u$:
$=\frac{1}{2}\int\cos^4 u\cos u\ du$
$=\frac{1}{2}\int(\cos^2 u)^2 \cos u\ du$
$=\frac{1}{2}\int(1-\sin^2 u)^2\cos u\ du$
Let $v=\sin u$. Then $dv=\cos u\ du$.
$=\frac{1}{2}\int(1-v^2)^2\ dv$
$=\frac{1}{2}\int(1-2v^2+v^4)\ dv$
$=\frac{1}{2}(v-\frac{2}{3}v^3+\frac{1}{5}v^5)$
$=\frac{1}{2}v-\frac{1}{3}v^3+\frac{1}{10}v^5$
$=\frac{1}{2}\sin u-\frac{1}{3}\sin^3 u+\frac{1}{10}\sin^5 u$
$=\boxed{\frac{1}{2}\sin(t^2)-\frac{1}{3}\sin^3(t^2)+\frac{1}{10}\sin^5(t^2)}$
