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 \[\operatorname{t} = si{{n}^{2}}u \Rightarrow dt = 2 sin u cos udu\] \[\Rightarrow \,\,\,dt=sin\,\,2\,udu\] \[\therefore \,\,{{I}_{1}}=\int_{0}^{x}{u\,\sin 2u\,du}\] Let \[{{\operatorname{I}}_{2}}\,=\,\,\int_{\,0}^{\,{{\cos }^{2}}x}{{{\cos }^{-1}}\,\sqrt{t} dt}\] Put \[\operatorname{t} = co{{s}^{2}}v \Rightarrow dt = -\,2cos v\,sin vdv\] \[\Rightarrow \,\,\,dt = -sin2v\,dv\] \[\therefore \,\,{{I}_{2}}= \int_{\frac{\pi }{2}}^{x}{v\left( -\,sin\,2v \right)dv}=\,\,-\int_{\,\frac{\pi }{2}}^{\,x}{\,v\,sin 2 vdv}\] \[= \,\,\int_{\,\frac{\pi }{2}}^{\,x}{ u sin 2udu}\] [change of variable] \[\therefore \,\,I={{I}_{1}}+{{I}_{2}}=\,\,\int_{\,0}^{\,x}{u\,\,\sin \,\,2\,\,udu}-\,\int_{\,\frac{\pi }{2}}^{\,x}{u\,sin2\,udu}\] \[=\,\,\,\int\limits_{0}^{\frac{\pi }{2}}{u\sin 2\,udu}+\,\,\int\limits_{\frac{\pi }{2}}^{x}{u\sin 2\,udu}-\,\int\limits_{\frac{\pi }{2}}^{x}{u\sin 2\,udu}\] [Integrate by parts]
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