Answer
The solution is $$\int_{0}^{\pi/2}x \sin2x dx=\frac{\pi}{4}.$$
Work Step by Step
To solve the integral $$\int_{0}^{\pi/2}x \sin2x dx$$ we will first use substitution $t=2x$ which gives us $x=\frac{t}{2},dx=\frac{dt}{2}$ and integration bounds become $t=0$ and $t=\pi$.
Putting all this into our integral we now have:
$$\int_{0}^{\pi/2}x \sin2x dx=\int_{0}^{\pi}\frac{t}{2} \sin t \frac{dt}{2}=\frac{1}{4}\int_{0}^{\pi}t\sin tdt.$$
Now we will integrate by parts using that $t=u$ and $\sin tdt=dv$ which gives us $dt=du$ and $-\cos t=v$. Putting all this into our integral we have:
$$\frac{1}{4}\int_{0}^{\pi}t\sin tdt=\frac{1}{4}\int_{0}^{\pi}udv=\frac{1}{4}\left(\left.uv\right|_{0}^{\pi}-\int_{0}^{\pi} vdu\right)=\frac{1}{4}(-\left.t\cos t\right|_{0}^{\pi})-\frac{1}{4}\int_{0}^{\pi}-\cos tdt=-\frac{1}{4}(\pi\cos \pi-0\cos0)+\frac{1}{4}\left.\sin t\right|_{0}^{\pi}=-\frac{\pi}{4}(-1)+\frac{1}{4}(\sin \pi-\sin0=\frac{\pi}{4}+\frac{1}{4}(0-0)=\frac{\pi}{4}.$$
