A) \[\frac{\pi }{2}-1\]
B) \[\frac{\pi }{2}+1\]
C) \[\frac{\pi }{2}\]
D) \[\pi +1\]
Correct Answer: A
Solution :
Let \[I=\int_{0}^{1}{\sqrt{\frac{1-x}{1+x}}}dx\] \[=\int_{0}^{1}{\sqrt{\frac{1-x}{1+x}\times \frac{1-x}{1-x}}}dx\] \[=\int_{0}^{1}{\frac{1-x}{\sqrt{1-{{x}^{2}}}}}dx\] \[=\int_{0}^{1}{\frac{1}{\sqrt{1-{{x}^{2}}}}}dx-\int_{0}^{1}{\frac{x}{\sqrt{1-{{x}^{2}}}}}dx\] Let \[1-{{x}^{2}}=t\]in second integral \[\Rightarrow \] \[-2xdx=dt\] \[\Rightarrow \] \[-\text{ }x\text{ }dx=\frac{1}{2}dt\] \[\therefore \] \[I=\int_{0}^{1}{\frac{1}{\sqrt{1-{{x}^{2}}}}dx}+\frac{1}{2}\int_{1}^{0}{\frac{1}{\sqrt{t}}}dt\] \[=[{{\sin }^{-1}}x]_{0}^{1}+\frac{1}{2}\left[ \frac{{{t}^{\frac{1}{2}}}}{\frac{1}{2}} \right]_{1}^{0}\] \[={{\sin }^{-1}}(1)-{{\sin }^{-1}}(0)-[0-1]\] \[I=\frac{\pi }{2}-1\]
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