Question

If ${\int }_{\frac{\pi }{3}}^x\sqrt{3-{\sin }^{2}t}dt+{\int }_0^y\cos tdt=0$ then evaluate $\frac{dy}{dx}.$

• A. $\frac{\sqrt{3-{\sin }^{2}x}}{\cos y}$
• B. $-\frac{\sqrt{3-{\sin }^{2}x}}{\cos y}$
• C. $\cos y\sqrt{3-{\sin }^{2}x}$
• D. $-\cos y\sqrt{3-{\sin }^{2}x}$

Answer 1

The correct option is B $-\frac{\sqrt{3-{\sin }^{2}x}}{\cos y}$

${\int }_{\pi /3}^x\sqrt{3-{\sin }^{2}t}dt+{\int }_0^y\cos tdt=0$

Differentiate with respect to $x$

$\sqrt{3-{\sin }^{2}x}.1+\sqrt{3-{\sin }^2\frac{\pi }{3}}\cdot 0+\cos y\frac{dy}{dx}=0$

$\sqrt{3-{\sin }^{2}x}+\cos y\frac{dy}{dx}=0$

$\cos y\frac{dy}{dx}=-\sqrt{3-{\sin }^{2}x}$

$\frac{dy}{dx}=-\frac{\sqrt{3-{\sin }^{2}x}}{\cos y}$