A) \[\frac{\pi }{4}\]
B) \[\frac{\pi }{2}\]
C) \[\pi \]
D) \[1\]
Correct Answer: D
Solution :
Let \[I=\int_{0}^{\pi /2}{x\sin x\,\,dx}\] \[=[-x\cos x+\int{\cos x\,\,dx]_{0}^{\pi /2}}\] \[=[-x\cos x+\sin x]_{0}^{\pi /2}\] \[=\left[ -\frac{\pi }{2}(0)+1-(0+0) \right]\] \[=1\]
You need to login to perform this action.
You will be redirected in
3 sec
