A) \[\frac{\pi }{2}\]
B) 1
C) \[\frac{\pi }{4}\]
D) None of these
Correct Answer: C
Solution :
We have \[I=\int_{0}^{{{\sin }^{2}}x}{{{\sin }^{-1}}\sqrt{t}\,dt+\int_{0}^{{{\cos }^{2}}x}{{{\cos }^{-1}}\sqrt{t}}\,dt}\] Putting \[t={{\sin }^{2}}u\]in the first integral and \[t={{\cos }^{2}}v\]in the second integral, we have \[I=\int_{0}^{x}{u\sin \,2u\,du\,-\int_{\pi /2}^{x}{v\sin 2v\,dv}}\] \[=\int_{0}^{\pi /2}{u\sin 2udu+\int_{\pi /2}^{x}{u\sin 2u\,du-\int_{\pi /2}^{x}{v\sin 2v\,dv}}}\] \[I=\int_{0}^{\pi /2}{u\sin 2udu=\left( \frac{-u\cos 2u}{2} \right)}_{0}^{\pi /2}+\frac{1}{2}\int_{0}^{\pi /2}{\cos 2u\,du}\] \[=\left( \frac{-u\cos 2u}{2} \right)_{0}^{\pi /2}+\frac{1}{4}(\sin 2u)_{0}^{\pi /2}=\frac{\pi }{4}\].
You need to login to perform this action.
You will be redirected in
3 sec
