Question

Evaluate${\int }_0^{{\sin }^{2}t}{\sin }^{-1}\sqrt{x}dx+{\int }_0^{{\cos }^{2}t}{\cos }^{-1}\sqrt{x}dx=k$, then $tan\left(k\right)=?$

Answer 1

Let $I={\int }_0^{{\sin }^{2}t}{\sin }^{-1}\sqrt{x}dx+{\int }_0^{{\cos }^{2}t}{\cos }^{-1}\sqrt{x}dx$
$I=I_1+I_2$
For $I_1$ put $\sqrt{x}=\sin \theta \Rightarrow x={\sin }^2\theta \Rightarrow dx=\sin 2\theta d\theta$
$\therefore I_1={\int }_0^t\theta \sin 2\theta d\theta$
And for $I_2$ put $\sqrt{x}=\cos \varphi$
$\therefore I_2={\int }_{\pi /2}^t-\varphi \sin 2\varphi d\varphi ={\int }_t^{\pi /2}\theta \sin 2\theta d\theta$
Hence
$I={\int }_0^t\theta \sin 2\theta d\theta +{\int }_t^{\pi /2}\theta \sin 2\theta d\theta ={\int }_0^{\pi /2}\theta \sin 2\theta d\theta$
$={[\theta (-\frac{\cos 2\theta }{2})-\left(1\right)(-\frac{\sin 2\theta }{4})]}_0^{\pi /2}=\frac{\pi }{2}.\frac{1}{2}+.0=\frac{\pi }{4}$