A) \[{{I}_{1}}={{I}_{2}}\]
B) \[{{I}_{2}}=\frac{\pi }{2}{{I}_{1}}\]
C) \[{{I}_{1}}+{{I}_{2}}=\frac{\pi }{2}x\]
D) \[{{I}_{1}}+{{I}_{2}}=\frac{\pi }{2}\]
E) \[{{I}_{1}}-{{I}_{2}}=\frac{\pi }{2}x\]
Correct Answer: C
Solution :
\[\because \] \[{{I}_{1}}=\int{{{\sin }^{-1}}x}dx\] And \[{{I}_{2}}=\int{{{\sin }^{-1}}}\sqrt{1-{{x}^{2}}}\,dx\] \[\Rightarrow \] \[{{I}_{2}}=\int{{{\cos }^{-1}}x\,dx}\] Now, \[{{I}_{1}}+{{I}_{2}}=\int{({{\sin }^{-1}}x+{{\cos }^{-1}}x)}dx\] \[=\int{\frac{\pi }{2}dx}=\frac{\pi }{2}x\] \[\therefore \] \[{{I}_{1}}+{{I}_{2}}=\frac{\pi }{2}x\]
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