Question

$y={\int }_{1/8}^{{\sin }^{2}x}{\sin }^{-1}\sqrt{t}dt+{\int }_{1/8}^{{\cos }^{2}x}{\cos }^{-1}\sqrt{t}dt$ , $0\le x\le x/2$.

• A. Is the equation of a straight line parallel to the $x$-axis
• B. Is the equation of a straight line which is the bisector of first quadrant
• C. Is the equation of a straight line which is the bisector of second quadrant
• D. None of the above

Answer 1

The correct option is A Is the equation of a straight line parallel to the $x$-axis


$\text{LEIBNIZ RULE}$

Here, we have to prove that,
$y$ is a constant or derivative of $y$ with respect to $x$ is zero.
$y={\int }_{1/8}^{{\sin }^{2}x}{\sin }^{-1}\sqrt{t}dt+{\int }_{1/8}^{{\cos }^{2}x}{\cos }^{-1}\sqrt{t}dt$ ....(i)
$\frac{dy}{dx}={\sin }^{-1}\sqrt{{\sin }^{2}x}\cdot 2\sin x\cdot \cos x+{\cos }^{-1}\sqrt{{\cos }^{2}x}$$(-2\sin x\cdot \cos x)=2x\sin x\cdot \cos x-2x\cdot \sin x\cdot \cos x$
$=0$ for all $x$.

$\therefore$ $y=constant$

Hence, the given equation is an equtaion of a line parallel to X axis