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