Answer
$\mathrm{i}+\mathrm{j}+\mathrm{k}$
Work Step by Step
Use the last boxed formula of this section:
$I= \displaystyle \int_{0}^{\pi/2}(3\sin^{2}t\cos t\mathrm{i}+3\sin t\cos^{2}t\mathrm{j}+2\sin t\cos t\mathrm{k})dt$
$=(\displaystyle \int_{0}^{\pi/2}3\sin^{2}t\cos tdt)\mathrm{i}+(\int_{0}^{\pi/2}3\sin t\cos^{2}tdt)\mathrm{j}+(\int_{0}^{\pi/2}2\sin t\cos tdt)\mathrm{k}$
First integral : $\displaystyle \left[\begin{array}{l}
u=\sin t\\
du=\cos tdt
\end{array}\right]=3\int_{0}^{\pi/2}u^{2}du=\left[u^{3}\right]_{0}^{\pi/2}$
Second integral: $\displaystyle \left[\begin{array}{l}
u=\cos t\\
du=-\sin tdt
\end{array}\right]=3\int_{0}^{\pi/2}u^{2}du=-\left[u^{3}\right]_{0}^{\pi/2}$
third integral: $\left[\begin{array}{l}
u=\sin t\\
du=\cos tdt
\end{array}\right]=$2$\displaystyle \int_{0}^{\pi/2}udu=\left[u^{2}\right]_{0}^{\pi/2}$
$I=[\sin^{3}t]_{0}^{\pi/2}\mathrm{i}+[-\cos^{3}t]_{0}^{\pi/2}\mathrm{j}+[\sin^{2}t]_{0}^{\pi/2}\mathrm{k}$
$=(1-0)\mathrm{i}+(0+1)\mathrm{j}+(1-0)\mathrm{k}$
$=\mathrm{i}+\mathrm{j}+\mathrm{k}$
