A) \[\pi -1\]
B) \[\frac{\pi }{2}\]
C) \[\frac{\pi -2}{2}\]
D) \[\frac{\pi }{4}\]
Correct Answer: C
Solution :
Let \[f(x)=\sin ({{\sin }^{-1}}[x])=\left\{ \begin{matrix} 0, & 0\le x<1 \\ 1, & 1\le x<\frac{\pi }{2} \\ \end{matrix} \right.\] \[\therefore I=\int\limits_{0}^{1}{0}dx+\int\limits_{1}^{\frac{\pi }{2}}{1dx=0}+\left( \frac{\pi }{2}+1 \right)=\frac{\pi -2}{2}\]You need to login to perform this action.
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