Question

${\int }_0^{\pi /2}x\sin xdx$ is equal to

• A. $\frac{4}{3}$
• B. $1$
• C. $0$
• D. $\frac{-8}{3}$

Answer 1

Answer: (B)

$1$

$I={\int }_0^{\pi /2}x\sin xdx$

$I=[-x\cos x{]}_0^{\pi /2}-{\int }_0^{\pi /2}1\times (-\cos x)dx$

$I=[-x\cos x{]}_0^{\pi /2}+[\sin x{]}_0^{\pi /2}$

$I=(-\frac{\pi }{2}\cos \frac{\pi }{2}+0\cos 0)+(\sin \frac{\pi }{2}-\sin 0)=1$