Question

The value of $\underset{0}{\overset{{\sin }^{2}x}{\int }}{\sin }^{-1}\sqrt{t}dt+\underset{0}{\overset{{\cos }^{2}x}{\int }}{\cos }^{-1}\sqrt{t}dt$ is

• A. $\frac{\pi }{2}$
• B. 1
• C. $\frac{\pi }{4}$
• D. none of these

Answer 1

Answer: (C)

Put $t=si{n}^{2}y$ in the integral
$\underset{0}{\overset{si{n}^{2}x}{\int }}si{n}^{-1}\sqrt{t}dt$
$\therefore dt=2sinydy=sin2ydy$
Put $t=co{s}^{2}u$ in the integral $\underset{0}{\overset{co{s}^{2}x}{\int }}co{s}^{-1}\sqrt{t}dt$
$\therefore dt=-2$ cos{\,} u sin{\,} u du = - sin{\,} 2u {\,}du
$\therefore$ the given result $=\underset{0}{\overset{x}{\int }}ysin2ydy-\underset{\pi /2}{\overset{x}{\int }}usin2udu$
$=\underset{0}{\overset{\pi /2}{\int }}ysin2ydy+\underset{\pi /2}{\overset{x}{\int }}ysin2ydy-\underset{\pi /2}{\overset{x}{\int }}usin2udu$
$=\underset{0}{\overset{\pi /2}{\int }}ysin2ydy$
$={\left|y\left(\frac{-cos2y}{2}\right)\right|}_0^{\pi /2}-\underset{0}{\overset{\pi /2}{\int }}1:-\frac{cos2y}{2}dy$
$=-\frac{\pi }{4}\left(-1\right)+\frac{1}{2}{\left|\frac{sin2y}{2}\right|}_0^{\pi /2}$
$=\frac{\pi }{4}$