Question

Evaluate $\underset{0}{\overset{{\sin }^{2}x}{\int }}{\sin }^{-1}\sqrt{t}td+\underset{0}{\overset{{\cos }^{2}x}{\int }}co{s}^{-1}\sqrt{t}dt\forall x$

Answer 1

Consider ${\int }_0^{{\sin }^{2}x}{\sin }^{-1}\sqrt{t}dt$

Let $t={\sin }^2\theta \Rightarrow dt=2\sin \theta \cos \theta d\theta =\sin 2\theta d\theta$

When $t=0\Rightarrow \theta =0$

When $t={\sin }^{2}x\Rightarrow {\sin }^2\theta ={\sin }^{2}x\Rightarrow \theta =x$

$\therefore {\int }_0^{{\sin }^{2}x}{\sin }^{-1}\sqrt{t}dt$

$={\int }_0^x{\sin }^{-1}\sqrt{{\sin }^2\theta }\sin 2\theta d\theta$

$={\int }_0^x{\sin }^{-1}\sin \theta \sin 2\theta d\theta$

$={\int }_0^x\theta \sin 2\theta d\theta$ ...........$\left(1\right)$

Consider ${\int }_0^{{\cos }^{2}x}{\cos }^{-1}\sqrt{t}dt$

Let $t={\cos }^2\theta \Rightarrow dt=-2\sin \theta \cos \theta d\theta =-\sin 2\theta d\theta$

When $t=0\Rightarrow \theta =\frac{\pi }{2}$

When $t={\cos }^{2}x\Rightarrow {\cos }^2\theta ={\cos }^{2}x\Rightarrow \theta =x$

$\therefore {\int }_{\frac{\pi }{2}}^x{\cos }^{-1}\sqrt{t}dt$

$=-{\int }_{\frac{\pi }{2}}^x{\cos }^{-1}\sqrt{{\cos }^2\theta }\sin 2\theta d\theta$

$=-{\int }_{\frac{\pi }{2}}^x{\cos }^{-1}\cos \theta \sin 2\theta d\theta$

$=-{\int }_{\frac{\pi }{2}}^x\theta \sin 2\theta d\theta$

$={\int }_x^{\frac{\pi }{2}}\theta \sin 2\theta d\theta$ ...........$\left(2\right)$

Using ${\int }_a^{b}f\left(x\right)dx+{\int }_b^{c}f\left(x\right)dx={\int }_a^{c}f\left(x\right)dx$ we add $\left(1\right)$ and $\left(2\right)$

${\int }_0^x\theta \sin 2\theta d\theta +{\int }_x^{\frac{\pi }{2}}\theta \sin 2\theta d\theta$

$={\int }_0^{\frac{\pi }{2}}\theta \sin 2\theta d\theta$

Let $u=\theta \Rightarrow du=d\theta$

$dv=\sin 2\theta \Rightarrow v=-\frac{\cos 2\theta }{2}$

$={[-\frac{\theta \cos 2\theta }{2}]}_0^{\frac{\pi }{2}}-\int -\frac{\cos 2\theta }{2}d\theta$

$=[\frac{\pi }{4}-0]-{\int }_0^{\frac{\pi }{2}}-\frac{\cos 2\theta }{2}d\theta$

$=\frac{\pi }{4}+\frac{1}{4}{\left[\sin 2\theta \right]}_0^{\frac{\pi }{2}}$

$=\frac{\pi }{4}+\frac{1}{4}{[0-0]}_0^{\frac{\pi }{2}}$

$=\frac{\pi }{4}$