Answer
$$\frac{\pi^2}{4}$$
Work Step by Step
Given $$\int_{0}^{\pi} t\cos^2tdt$$
Since
\begin{align*}
\int_{0}^{\pi} t\cos^2tdt&=\frac{1}{2}\int_{0}^{\pi} t[1+\cos2t]dt\\
&=\frac{1}{2}\int_{0}^{\pi} tdt+\frac{1}{2}\int_{0}^{\pi}t\cos2t dt\\
&=\frac{1}{4} t^2 \bigg|_{0}^{\pi}+\frac{1}{2}\int_{0}^{\pi}t\cos2t dt\\
\end{align*}
To evaluate $\displaystyle\int_{0}^{\pi}t\cos2t dt$ Let
\begin{align*}
u&=t \ \ \ \ \ \ \ \ \ \ \ \ dv=\cos2t dt\\
du&=dt \ \ \ \ \ \ \ \ \ \ \ \ v=\frac{1}{2}\sin2t
\end{align*}
Then
\begin{align*}
\int_{0}^{\pi}t\cos2t dt&=\frac{1}{2}t\sin2t \bigg|_{0}^{\pi} -\frac{1}{2}\int_{0}^{\pi}\sin2t dt\\
&=\frac{1}{4} \cos 2t\bigg|_{0}^{\pi}\\
&=0
\end{align*}
Hence
$$\int_{0}^{\pi} t\cos^2tdt=\frac{\pi^2}{4} $$