Question

If $\int \frac{si{n}^{-1}\sqrt{x}-co{s}^{-1}\sqrt{x}}{si{n}^{-1}\sqrt{x}+co{s}^{-1}\sqrt{x}}dx=\frac{A}{748\pi }[-(si{n}^{-1}\sqrt{x}\left)\right(1-2x)+\sqrt{x}\sqrt{1-x}]-x+C$ then A is equal to

Answer 1

$I=\int \frac{{\sin }^{-1}\sqrt{x}-{\cos }^{-1}\sqrt{x}}{{\sin }^{-1}\sqrt{x}+{\cos }^{-1}\sqrt{x}}dx$
Substitute ${\sin }^{-1}\sqrt{x}=t\Rightarrow x={\sin }^{2}t\Rightarrow dx=\sin 2tdt$
$I=\int \frac{t-(\frac{\pi }{2}-t)\left(\sin 2t\right)}{\frac{\pi }{2}}dt=\frac{2}{\pi }\int (2t-\frac{\pi }{2})\left(\sin 2t\right)dt=\frac{2}{\pi }\int 2t\sin 2tdt-\int \sin 2tdt=\frac{2}{\pi }(-t\cos 2t+\int \cos 2tdt)-\int \sin 2tdt=-\frac{2t}{\pi }\cos 2t+\frac{\sin 2t}{\pi }+\frac{\cos 2t}{2}+c=\frac{2}{\pi }(-{\sin }^{-1}\sqrt{x}(1-2x)+\sqrt{x}\sqrt{1-x})-x+c$
Therefore $A=1496$