Answer
$\displaystyle \frac{\pi^{2}}{4}$
Work Step by Step
Strategy: simplify $\cos^{2}t$ using double angle identity.
$\displaystyle \int_{0}^{\pi}t\cos^{2}tdt=\int_{0}^{\pi}t\cdot\frac{1+\cos 2t}{2}dt$
$=\displaystyle \frac{1}{2}\int_{0}^{\pi}tdt+\frac{1}{2}\int_{0}^{\pi}t\cos 2tdt$
First integral : table: 1. $\displaystyle \int x^{\mathrm{n}}dx=\frac{x^{n+1}}{n+1} (n\neq-1)$,
Second integral: $\displaystyle \int udv$ by parts, $\left[\begin{array}{ll}
u=t & dv=\mathrm{c}\mathrm{o}\mathrm{s}2tdt\\
du=dt & v=\frac{1}{2}\mathrm{s}\mathrm{i}\mathrm{n}2t
\end{array}\right]$
$=\displaystyle \frac{1}{2}[\frac{1}{2}t^{2}]_{0}^{\pi}+\frac{1}{2}[uv]_{0}^{\pi}-\frac{1}{2}\int_{0}^{\pi}vdudt$
$=\displaystyle \frac{1}{2}[\frac{1}{2}t^{2}]_{0}^{\pi}+\frac{1}{2}[\frac{1}{2}t\sin 2t]_{0}^{\pi}-\frac{1}{2}\int_{0}^{\pi}\frac{1}{2}\sin 2tdt$
$=\displaystyle \frac{\pi^{2}}{4}+0-\frac{1}{4}[-\frac{1}{2}\cos 2t]_{0}^{\pi}$
$=\displaystyle \frac{\pi^{2}}{4}+\frac{1}{8}(1-1)$
$=\displaystyle \frac{\pi^{2}}{4}$
