A) \[0\]
B) \[1\]
C) \[\frac{\pi }{4}\]
D) \[\frac{{{\pi }^{2}}}{2}\]
E) \[\frac{\pi }{2}\]
Correct Answer: C
Solution :
Let\[I=\int_{0}^{1}{\frac{x\,dx}{[x+\sqrt{1-{{x}^{2}}}]\sqrt{1-{{x}^{2}}}}}\] Put \[x=\sin \theta \Rightarrow dx=\cos \theta \,d\theta \] \[\therefore \] \[I=\int_{0}^{\pi /2}{\frac{\sin \theta .\cos \theta d\theta }{(\sin \theta +\cos \theta ).\cos \theta }}\] \[\Rightarrow \]\[I=\int_{0}^{\pi /2}{\frac{\sin \theta }{\sin \theta +\cos \theta }}d\theta \] ?. (i) \[\Rightarrow \]\[I=\int_{0}^{\pi /2}{\frac{\sin (\pi /2-\theta )}{\sin (\pi /2-\theta )+\cos (\pi /2-\theta )}}d\theta \] \[\Rightarrow \]\[I=\int_{0}^{\pi /2}{\frac{\cos \theta }{\sin \theta +\cos \theta }}d\theta \] ?. (ii) On adding Eqs (i) and (ii), we get \[2I=\int_{0}^{\pi /2}{\left( \frac{\sin \theta +\cos \theta }{\sin \theta +\cos \theta } \right)}d\theta =\int_{0}^{\pi /2}{1\,d\theta =\frac{\pi }{2}}\] \[\Rightarrow \] \[I=\frac{\pi }{4}\]
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