Question

If $y=\underset{1/8}{\overset{{\sin }^{2}x}{\int }}{\sin }^{-1}\sqrt{t}dt+{\int }_{1/8}^{{\cos }^{2}x}{\cos }^{-1}\sqrt{t}dt,$ where $0\le x\le \pi /2,$ then which of the following is correct

• A. $y=\frac{3\pi }{4}$
• B. $y=\frac{3\pi }{2}$
• C. $y=\frac{3\pi }{16}$
• D. $y=\frac{3\pi }{8}$

Answer 1

The correct option is C $y=\frac{3\pi }{16}$

$y=\underset{1/8}{\overset{{\sin }^{2}x}{\int }}{\sin }^{-1}\sqrt{t}dt+\underset{1/8}{\overset{{\cos }^{2}x}{\int }}{\cos }^{-1}\sqrt{t}dt...\left(1\right)$
$\frac{dy}{dx}={\sin }^{-1}\sqrt{{\sin }^{2}x}.2\sin x\cos x+{\cos }^{-1}\sqrt{{\cos }^{2}x}.(-2\cos x\sin x)$
$=2x\sin x\cos x-2x\sin x\cos x$
$\Rightarrow \frac{dy}{dx}=0\forall x\in [0,\frac{\pi }{2}]$
$\therefore$ The curve in equation $\left(1\right)$ is a straight line parallel to the $x-axis.$
Now, since y is constant, it is independent of $x.$
So let's select any $x$, say $x=\frac{\pi }{4}$ (convenient value)
$y=\underset{1/8}{\overset{1/2}{\int }}{\sin }^{-1}\sqrt{t}dt+\underset{1/8}{\overset{1/2}{\int }}{\cos }^{-1}\sqrt{t}dt=\underset{1/8}{\overset{1/2}{\int }}({\sin }^{-1}\sqrt{t}+{\cos }^{-1}\sqrt{t})dt=\underset{1/8}{\overset{1/2}{\int }}\frac{\pi }{2}dt=\frac{\pi }{2}[\frac{1}{2}-\frac{1}{8}]=\frac{3\pi }{16}$
Therefore, equation of the line is $y=\frac{3\pi }{16}.$