A) \[\left( \frac{\pi }{2}-1 \right)\]
B) \[\left( \frac{\pi }{2}+1 \right)\]
C) \[\frac{\pi }{2}\]
D) \[(\pi +1)\]
Correct Answer: A
Solution :
\[I=\int_{0}^{1}{\sqrt{\frac{1-x}{1+x}}dx=\int_{0}^{1}{\sqrt{\frac{1-x}{1+x}}}.\frac{\sqrt{1-x}}{\sqrt{1-x}}dx}\] \[=\int_{0}^{1}{\frac{1-x}{\sqrt{1-{{x}^{2}}}}dx=\int_{0}^{1}{\frac{dx}{\sqrt{1-{{x}^{2}}}}}-\int_{0}^{1}{\frac{x}{\sqrt{1-{{x}^{2}}}}\,}dx}\] \[I=[{{\sin }^{-1}}x]_{0}^{1}+[\sqrt{1-{{x}^{2}}}]_{0}^{1}=\frac{\pi }{2}-1\].
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