Question

Evaluate the following definite integrals:

${\int }_0^{\frac{\mathrm{\pi }}{2}}{x}^2\sin xdx$ [CBSE 2014]

Answer 1

$\mathrm{Let}I={\int }_0^{\frac{\mathrm{\pi }}{2}}{x}^2\sin x$

Applying integration by parts, we have

$$\begin{gathered} I=x^2\left(-\cos x\right){|}_0^{\frac{\mathrm{\pi }}{2}}-{\int }_0^{\frac{\mathrm{\pi }}{2}}2x\left(-\cos x\right)dx \\ \Rightarrow I=\left(0-0\right)+2{\int }_0^{\frac{\mathrm{\pi }}{2}}x\cos xdx\left(\cos \frac{\mathrm{\pi }}{2}=0\right) \end{gathered}$$

Again applying integration by parts, we have

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