\[{{I}_{1}}=\int{{{\sin }^{-1}}x\,\,dx}\] and \[{{I}_{2}}=\int{{{\sin }^{-1}}\sqrt{1-{{x}^{2}}}}dx\]then [Kerala (Engg.) 2005]

\[{{I}_{1}}+{{I}_{2}}=\pi /2x\]

\[{{I}_{1}}-{{I}_{2}}=\pi /2x\]

\[{{I}_{1}}=\int{{{\sin }^{-1}}x\,\,dx}\] and \[{{I}_{2}}=\int{{{\sin

1.	\[{{I}_{1}}=\int{{{\sin }^{-1}}x\,\,dx}\] and \[{{I}_{2}}=\int{{{\sin }^{-1}}\sqrt{1-{{x}^{2}}}}dx\]then [Kerala (Engg.) 2005]
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.	\[{{I}_{1}}-{{I}_{2}}=\pi /2x\]
Answer» D. \[{{I}_{1}}+{{I}_{2}}=\pi /2\]

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\[{{I}_{1}}=\int{{{\sin }^{-1}}x\,\,dx}\] and \[{{I}_{2}}=\int{{{\sin }^{-1}}\sqrt{1-{{x}^{2}}}}dx\]then [Kerala (Engg.) 2005]

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