A) \[{{\sin }^{-1}}x+\sqrt{1-{{x}^{2}}}+C\]
B) \[{{\sin }^{-1}}x-2\sqrt{1-{{x}^{2}}}+C\]
C) \[2{{\sin }^{-1}}x-\sqrt{1-{{x}^{2}}}+C\]
D) \[{{\sin }^{-1}}x-\sqrt{1-{{x}^{2}}}+C\]
E) \[-{{\cos }^{-1}}x-\sqrt{1-{{x}^{2}}}+C\]
Correct Answer: A
Solution :
\[\int{\sqrt{\frac{1-x}{1+x}}}dx\int{\sqrt{\frac{(1-x)(1-x)}{(1+x)(1-x)}}}dx\] After rationalizing \[=\int{\frac{(1-x)}{\sqrt{1-{{x}^{2}}}}}dx\] \[=\int{\frac{dx}{\sqrt{1-{{x}^{2}}}}-\frac{1}{2}\int{\frac{2x}{\sqrt{1-{{x}^{2}}}}}}dx\] \[={{\sin }^{-1}}x-\frac{1}{2}(-2\sqrt{1-{{x}^{2}}})+C\] \[={{\sin }^{-1}}x+\sqrt{1+{{x}^{2}}}+C\]
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