Question

Solve the equation:-
If ${\int }_{\frac{\pi }{3}}^x\sqrt{3-co{s}^{2}t}dt+{\int }_0^{y^2}sintdt=0$, then $\frac{dy}{dx}$=

Answer 1

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

${\int }_{\frac{\pi }{3}}^x\sqrt{3-{\cos }^{2}t}dt+{[-\cos t]}_0^{y^2}=0$

$\Rightarrow {\int }_{\frac{\pi }{3}}^x\sqrt{3-{\cos }^{2}t}dt+[-\cos y^2+1]=0\Rightarrow {\int }_{\frac{\pi }{3}}^x\sqrt{3-{\cos }^{2}t}dt=\cos y^2-1$

Differentiating both sides w r t $x$

$\sqrt{3-{\cos }^{2}x}.\frac{dx}{dx}-\sqrt{3-{\cos }^2\frac{\pi }{3}}.\frac{d}{dx}\left(\frac{\pi }{3}\right)=\frac{d}{dx}(\cos y^2-1)$

$\Rightarrow \sqrt{3-{\cos }^{2}x}-\sqrt{3-{\cos }^2\frac{\pi }{3}}.0=-\sin y^2.2y.\frac{dy}{dx}$

$\Rightarrow \sqrt{3-{\cos }^{2}x}=-\sin y^2.2y.\frac{dy}{dx}$

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