A) \[\pi \]
B) \[\frac{\pi }{2}\]
C) \[\frac{\pi }{4}\]
D) 1
Correct Answer: C
Solution :
| [c] Let \[{{I}_{1}}=\int_{0}^{{{\sin }^{2}}x}{{{\sin }^{-1}}\sqrt{t}\,\,dt}\] |
| Put \[t={{\sin }^{2}}u\Rightarrow dt=2\sin \,\,u\,\,\cos \,\,udu\] |
| \[\Rightarrow dt=\sin \,\,2\,\,udu\] |
| \[\therefore \,\,\,\,\,{{I}_{1}}=\int_{0}^{x}{u\,\,\sin \,\,2udu}\] |
| Let \[{{I}_{2}}=\int_{0}^{{{\cos }^{2}}x}{{{\cos }^{-1}}\sqrt{t}\,dt}\] |
| Put \[t={{\cos }^{2}}v\Rightarrow dt=-2\cos \,\,v\,\,sin\,\,vdv\] |
| \[\Rightarrow dt=-\sin \,\,2\,vdv\] |
| \[\therefore {{I}_{2}}=\int_{\frac{\pi }{2}}^{x}{v}(-sin2v)dv=-\int_{\frac{\pi }{2}}^{x}{v\sin 2\,vdv}\] |
| \[=-\int_{\frac{\pi }{2}}^{x}{u\,\,\sin \,\,2udu}\] [change of variable] |
| \[\therefore I={{I}_{1}}+{{I}_{2}}=\int_{0}^{x}{u\sin \,\,2udu-\int_{\frac{\pi }{2}}^{x}{u\sin \,\,2\,udu}}\] |
| \[=\int\limits_{0}^{\frac{\pi }{2}}{u\,\,\sin \,\,2udu+\int\limits_{\frac{\pi }{2}}^{x}{u\,\,\sin \,\,2udu-\int\limits_{\frac{\pi }{2}}^{x}{u\,\,\sin \,\,2\,\,udu}}}\]\[=\int\limits_{0}^{\frac{\pi }{2}}{u\sin 2\,\,udu}=\frac{\pi }{4}\] [Integrate by parts] |
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