Question

Evaluate:${\int }_0^{\frac{\mathrm{\pi }}{2}}x\sin xdx$

• A. $0$
• B. $1$
• C. $-1$
• D. $2$

Answer 1

The correct option is B

$1$

Explanation for Correct answer:

Let,$u=x$

on differentiating with respect to $x$, we get,

$du=dx$

Let, $v=-\cos x$

on differentiating with respect to $x$,

$dv=\sin xdx$

$$\begin{gathered} \int x\sin xdx=\left[x(-\cos x)\right]-\int -\cos x.dx\left[\because \int udv=uv-\int v.du\right] \\ =\left[x(-\cos x)\right]+\left[\sin x\right]\left[\because \int \cos xdx=\sin x\right] \\ {\int }_0^{\frac{\mathrm{\pi }}{2}}x\sin xdx={\left[x(-\cos x)+\sin x\right]}_0^{\frac{\mathrm{\pi }}{2}} \\ =\left[\frac{\mathrm{\pi }}{2}(-\cos \frac{\mathrm{\pi }}{2})+\sin \frac{\mathrm{\pi }}{2}\right]-\left[0(-\cos 0)+\sin 0\right]\left[\because \cos \frac{\mathrm{\pi }}{2}=0,\sin \frac{\mathrm{\pi }}{2}=1\right] \\ =1-0 \\ =1 \end{gathered}$$

Hence, option (B) is correct.