Question

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

Answer 1

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

differentiating on both sides

$\frac{d}{dx}{\int }_{a\left(x\right)}^{b\left(x\right)}f(x,t)dt={\int }_{a\left(x\right)}^{b\left(x\right)}\frac{\partial f}{\partial x}dt+f(x,b(x\left)\right)\frac{d\left(b\right(x\left)\right)}{dx}-f(x,a(x\left)\right)\frac{d\left(a\right(x\left)\right)}{dx}$

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