A) \[{{I}_{1}}={{I}_{2}}\]
B) \[{{I}_{2}}=\pi /2{{I}_{1}}\]
C) \[{{I}_{1}}+{{I}_{2}}=\pi /2x\]
D) \[{{I}_{1}}+{{I}_{2}}=\pi /2\]
E) \[{{I}_{1}}-{{I}_{2}}=\pi /2x\]
Correct Answer: C
Solution :
\[{{I}_{1}}=\int{{{\sin }^{-1}}xdx}\] Let \[{{\sin }^{-1}}x=\theta \]Þ \[x=\sin \theta \] Þ \[dx=\cos \theta \,d\theta \] \[{{I}_{1}}=\int{\theta \cos \theta d\theta }\]\[=\theta \sin \theta -\int{\sin \theta d\theta }\]\[=\theta \sin \theta +\cos \theta \] \[=x{{\sin }^{-1}}x+\sqrt{1-{{x}^{2}}}\] \[{{I}_{2}}=\int{{{\sin }^{-1}}\sqrt{1-{{x}^{2}}}}dx\]\[=\int{{{\cos }^{-1}}xdx}\] Let \[\cos \varphi =x,\] Hence \[-\sin \varphi \,d\varphi =dx\] \[{{I}_{2}}=-\int{\varphi \sin \varphi d\varphi }\]\[=\varphi \cos \varphi +\int{-\cos \varphi d\varphi }\] \[=\varphi \cos \varphi -\sin \varphi \]\[=x{{\cos }^{-1}}x-\sqrt{1-{{x}^{2}}}\] \[{{I}_{1}}+{{I}_{2}}=x({{\cos }^{-1}}x+{{\sin }^{-1}}x)=\frac{\pi }{2}x\].
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