A) \[\pi \]
B) \[\frac{\pi }{2}\]
C) \[\frac{\pi }{4}\]
D) 1
Correct Answer: C
Solution :
Let\[{{I}_{1}}=\int_{0}^{{{\sin }^{2}}x}{{{\sin }^{-1}}}\sqrt{t}dt\] Put\[t={{\sin }^{2}}u\Rightarrow dt=2\sin u\cos udu\] \[\Rightarrow \]\[dt=\sin 2udu\] \[\therefore \]\[{{I}_{1}}=\int_{0}^{x}{u}\sin 2udu\] Let\[{{I}_{2}}=\int_{0}^{{{\cos }^{2}}x}{{{\cos }^{-1}}}\sqrt{t}dt\] Put\[t={{\cos }^{2}}v\Rightarrow dt=-2\cos v\sin vdv\] \[\Rightarrow \]\[dt=-\sin 2vdv\] \[\therefore \]\[{{I}_{2}}=\int_{\frac{\pi }{2}}^{x}{v(-\sin 2v)dv=-\int_{\frac{\pi }{2}}^{x}{v\sin 2vdv}}\] \[=-\int_{\frac{\pi }{2}}^{x}{u\sin 2udv}\][change of variable] \[\therefore \]\[I={{I}_{1}}+{{I}_{2}}=\int_{0}^{x}{u\sin 2udv}-\int_{\frac{\pi }{2}}^{x}{u\sin 2udu}\] \[\int\limits_{0}^{\frac{\pi }{2}}{u\sin 2udu+}\int\limits_{\frac{\pi }{2}}^{x}{u\sin 2udu-}\int\limits_{\frac{\pi }{2}}^{x}{u\sin 2udu}\] \[=\int\limits_{0}^{\frac{\pi }{2}}{u\sin 2udu}=\frac{\pi }{4}\][Integrate by parts]
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