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) \[\frac{2}{\pi }[\sqrt{x-{{x}^{2}}}+(1-2x){{\sin }^{-1}}\sqrt{x}]-x+C\]
D) None of the above
Correct Answer: B
Solution :
\[I=\int{\frac{{{\sin }^{-1}}\sqrt{x}-(\pi /2-{{\sin }^{-1}}\sqrt{x})}{\pi /2}}dx\] \[\{{{\sin }^{-1}}\sqrt{x}+{{\cos }^{-1}}\sqrt{x}=\pi /2\}\] \[=4/\pi \int{{{\sin }^{-1}}\sqrt{x}dx}-\int{1}\,dx\] \[=4/\pi \int{{{\sin }^{-1}}\sqrt{x}dx}-x+C\] Apply integration by parts \[=\frac{4}{\pi }\left\{ {{\sin }^{-1}}\sqrt{x}.x-\int{\frac{1}{\sqrt{1-x}}.\frac{1}{2\sqrt{x}}.xdx} \right\}\] \[-x+C\] \[=\frac{4}{\pi }\left\{ x{{\sin }^{-1}}\sqrt{x}-\frac{1}{2}\int{\sqrt{\frac{x}{1-x}}dx} \right\}-x+C\] Put \[x=si{{n}^{2}}\theta \] \[dx=2\sin \theta .\cos \theta \,d\theta \] \[=\frac{4}{\pi }\left\{ x{{\sin }^{-1}}\sqrt{x}-\frac{1}{2}\int{\frac{\sin \theta }{\cos \theta }}.2\sin \theta .\cos \theta .d\theta \right\}\] \[-x+C\] \[=\frac{4}{x}\{x{{\sin }^{-1}}\sqrt{x}-\int{{{\sin }^{2}}\theta d\theta \}-x+C}\] \[=\frac{4}{x}\left\{ x{{\sin }^{-1}}\sqrt{x}-\frac{1}{2}\int{(1-\cos 2\theta )d\theta } \right\}-x+C\] \[=\frac{4}{x}\left\{ x{{\sin }^{-1}}\sqrt{x}-\frac{1}{2}\left( \theta -\frac{\sin 2\theta }{2} \right) \right\}-x+C\] \[=\frac{4}{x}\left\{ x{{\sin }^{-1}}\sqrt{x}-\frac{\theta }{2}\frac{\sin \theta .\cos \theta }{2} \right\}-x+C\] \[=\frac{4}{x}\left\{ x{{\sin }^{-1}}\sqrt{x}-\frac{1}{2}{{\sin }^{-1}}\sqrt{x}+\frac{1}{2}\sqrt{x}\sqrt{1-x} \right\}\] \[-x+C\] \[=\frac{4}{x}\left\{ \frac{1}{2}\sqrt{x-{{x}^{2}}}+\frac{(2x-1)}{2}.{{\sin }^{-1}}\sqrt{x} \right\}-x+C\] \[=\frac{2}{x}\{\sqrt{x-{{x}^{2}}}-(1-2x){{\sin }^{-1}}\sqrt{x}\}-x+C\]
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