A) \[\frac{2}{\pi }\{\sqrt{x-{{x}^{2}}}\}-(1+2x){{\sin }^{-1}}\sqrt{x}\}-x+c\]
B) \[\frac{2}{\pi }\{\sqrt{x-{{x}^{2}}}-(1-2x){{\sin }^{-1}}\sqrt{x}\}-x+c\]
C) \[\frac{2}{\pi }\{\sqrt{x-{{x}^{2}}}-(1-2x){{\sin }^{-1}}x\}-x+c\]
D) None of these
Correct Answer: B
Solution :
Let \[I=\int_{{}}^{{}}{\frac{{{\sin }^{-1}}\sqrt{x}-{{\cos }^{-1}}\sqrt{x}}{{{\sin }^{-1}}\sqrt{x}+{{\cos }^{-1}}\sqrt{x}}dx}\] \[=\int_{{}}^{{}}{\frac{{{\sin }^{-1}}\sqrt{x}-\left( \frac{\pi }{2}-{{\sin }^{-1}}\sqrt{x} \right)}{\frac{\pi }{2}}dx}\] \[=\frac{2}{\pi }\int_{{}}^{{}}{\left\{ {{\sin }^{-1}}\sqrt{x}-\frac{\pi }{2} \right\}dx}\] \[=\frac{4}{\pi }\int_{{}}^{{}}{{{\sin }^{-1}}\sqrt{x}dx}-x-c\] ?(i) Now,\[\int_{{}}^{{}}{{{\sin }^{-1}}\sqrt{x}}dx\] \[=\int_{{}}^{{}}{\theta .\sin 2\theta d\theta ,}\] where \[x={{\sin }^{2}}\theta \] \[\Rightarrow \]\[dx=\sin 2\theta d\theta \] \[=-\frac{\theta \cos 2\theta }{2}+\int_{{}}^{{}}{\frac{1}{2}\cos 2\theta d\theta }\] \[=-\frac{\theta }{2}\cos 2\theta +\frac{1}{4}\sin 2\theta \] \[=-\frac{1}{2}\theta (1-2{{\sin }^{2}}\theta )+\frac{1}{2}\sin \theta \sqrt{1-{{\sin }^{2}}\theta }\] \[=-\frac{1}{2}{{\sin }^{-1}}\sqrt{x}(1-2x)+\frac{1}{2}\sqrt{x}\sqrt{1-x}\]?(ii) From Eqs. (i) and (ii), we get \[I=\frac{\pi }{\pi }\left\{ -\frac{1}{2}(1-2x){{\sin }^{-1}}\sqrt{x}+\frac{1}{2}\sqrt{x-{{x}^{2}}} \right\}\] \[-x+c\] \[=\frac{\pi }{\pi }\{\sqrt{x-{{x}^{2}}}-(1-2x){{\sin }^{-1}}\sqrt{x}\}-x+c\]
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