Question

For $0\le x\le \frac{\pi }{2}$, the value of ${\int }_0^{si{n}^{2}x}si{n}^{-1}\sqrt{t}dt+{\int }_0^{co{s}^{2}x}co{s}^{-1}\sqrt{t}dt$ equals

• A. $-\frac{\pi }{4}$
• B. $0$
• C. $1$
• D. $\frac{\pi }{4}$

Answer 1

The correct option is D $\frac{\pi }{4}$

Let $I_1={\int }_0^{si{n}^{2}x}si{n}^{-1}\sqrt{t}dt,put\theta =si{n}^{-1}\sqrt{t}$
$\Rightarrow t=si{n}^2\theta$
$dt=sin2\theta d\theta$
$whent=si{n}^{2}x,\theta =x$
$whent=0,\theta =0$
$I_2={\int }_0^x\theta sin2\theta d\theta$
$I_2={\int }_0^{co{s}^{2}x}co{s}^{-1}\sqrt{t}dt,puty=co{s}^{-1}\sqrt{t}$
$\Rightarrow t=co{s}^{2}y$
$dt=-sin2ydy$
$whent=co{s}^{2}x,y=x$
$t=0,y=\frac{\pi }{2}$
$\therefore I_2={\int }_{\frac{\pi }{2}}^{x}y(-sin2ydy)={\int }_x^{\frac{\pi }{2}}ysin2ydy={\int }_x^{\frac{\pi }{2}}\theta sin2\theta d\theta$
$\therefore I_1+I_2={\int }_0^{\frac{\pi }{2}}\theta sin2\theta$ $d\theta =\frac{\pi }{4}$