question_answer

Evaluate the integral \[\int{\frac{{{\sin }^{-1}}\sqrt{x}-{{\cos }^{-1}}\sqrt{x}}{{{\sin }^{-1}}\sqrt{x}+{{\cos }^{-1}}\sqrt{x}}\,dx.}\]

Answer:

Let        \[\left[ xy\cos \,\left( \frac{y}{x} \right)+{{y}^{2}}\sin \,\left( \frac{y}{x} \right) \right]dy\] \[=\int{\frac{{{\sin }^{-\,1}}\sqrt{x}-\left( \frac{\pi }{2}-{{\sin }^{-\,1}}\sqrt{x} \right)}{\frac{\pi }{2}}}\,\,dx\] \[\left[ \because {{\sin }^{-\,1}}\sqrt{x}+{{\cos }^{-\,1}}\sqrt{x}=\frac{\pi }{2} \right]\] \[=\frac{2}{\pi }\int{\left( 2{{\sin }^{-\,1}}\sqrt{x}-\frac{\pi }{2} \right)\,\,dx}\] \[=\frac{4}{\pi }\int{{{\sin }^{-\,1}}\sqrt{x}}dx-\int{1\cdot dx}\] \[=\frac{4}{\pi }\int{{{\sin }^{-\,1}}\sqrt{x}}dx-x+C\] \[=\frac{4}{\pi }{{l}_{1}}-x+C,\] Where   \[{{l}_{1}}=\int{{{\sin }^{-\,1}}}\sqrt{x}\,dx\]                 ?(i) Put\[x={{\sin }^{2}}\theta \Rightarrow dx=2\sin \theta \cos \theta d\theta =\sin 2\theta d\theta ,\] \[\therefore \]\[{{l}_{1}}=\int{\theta \sin 2\theta d\theta =-\,\theta \frac{\cos 2\theta }{2}+\int{\frac{1}{2}\cos 2\theta d\theta }}\] \[=-\frac{\theta }{2}\cos 2\theta +\frac{1}{4}\sin 2\theta \] \[=\frac{-\,1}{2}\theta \,(1-2{{\sin }^{2}}\theta )+\frac{1}{2}\sin \theta \sqrt{1-{{\sin }^{2}}}\theta \] \[=\frac{-\,1}{2}(1-2x)\,{{\sin }^{-\,1}}\sqrt{x}+\frac{1}{2}\sqrt{x}\sqrt{1-x}\]     ?(ii) From Eqs. (i) and (ii), we get \[{{l}_{1}}=\frac{4}{\pi }\left\{ \frac{-\,1}{2}(1-2x)\,{{\sin }^{-\,1}}\sqrt{x}+\frac{1}{2}\sqrt{x-{{x}^{2}}} \right\}-x+C\] \[=\frac{2}{\pi }\{\sqrt{x-{{x}^{2}}}-(1-2x)\,{{\sin }^{-\,1}}\sqrt{x}\}-x+C\]