Question

$\underset{0}{\overset{\mathrm{\pi }/2}{\int }}x\sin xdx$ is equal to

(a) π/4
(b) π/2
(c) π
(d) 1

Answer 1

(d) 1

$$\begin{gathered} \mathrm{We}\mathrm{have}, \\ I={\int }_0^{\frac{\mathrm{\pi }}{2}}x\sin xdx \\ ={\left[-x\cos x\right]}_0^{\frac{\mathrm{\pi }}{2}}-{\int }_0^{\frac{\mathrm{\pi }}{2}}1\left(-\cos x\right)dx \\ ={\left[-x\cos x\right]}_0^{\frac{\mathrm{\pi }}{2}}+{\int }_0^{\frac{\mathrm{\pi }}{2}}\cos xdx \\ =-{\left[x\cos x\right]}_0^{\frac{\mathrm{\pi }}{2}}+{\left[\sin x\right]}_0^{\frac{\mathrm{\pi }}{2}} \\ =-\left[0-0\right]+\left[1-0\right] \\ =1 \end{gathered}$$