A) \[-2\sqrt{\frac{1+\sqrt{x}}{1-\sqrt{x}}}+C\]
B) \[-2\sqrt{\frac{1-\sqrt{x}}{1+\sqrt{x}}}+C\]
C) \[-\sqrt{\frac{1-\sqrt{x}}{1+\sqrt{x}}}+C\]
D) \[2\sqrt{\frac{1+\sqrt{x}}{1-\sqrt{x}}}+C\]
Correct Answer: B
Solution :
\[I=\int{\frac{dx}{\left( 1+\sqrt{x} \right)\sqrt{x-{{x}^{2}}}}}\] put \[x={{\cos }^{2}}\theta \] \[dx=-2\cos \theta \,\sin \theta \,d\theta \] \[I=\int{\frac{-2\sin \theta \cos \theta d\theta }{(1+\cos \theta )\cos \theta \sin \theta }}\]\[=-2\int{\frac{d\theta }{2{{\cos }^{2}}\theta /2}}\] \[=-\int{{{\sec }^{2}}\left( \frac{\theta }{2} \right)d\theta }\] \[\therefore \,\cos \theta =\sqrt{x}\] \[=-2\tan \theta /2+C\] \[\frac{1-{{\tan }^{2}}\theta /2}{1+{{\tan }^{2}}\theta /2}=\sqrt{x}\] \[=2\sqrt{\frac{1-\sqrt{x}}{1+\sqrt{x}}}+c\] \[\Rightarrow \] \[{{\tan }^{2}}\left( \frac{\theta }{2} \right)=\frac{1-\sqrt{x}}{1+\sqrt{x}}\]
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