If int_{pi /2}^x sqrt{3 - 2sin^2 u} ,du + int | vedclass

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Question

(B) Given the equation: $\int_{\pi /2}^x \sqrt{3 - 2\sin^2 u} \,du + \int_0^y \cos t \,dt = 0.$ Applying the Leibniz rule for differentiation under th

Vedclass · Accepted Answer

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

Vedclass · Answer

(B) Given the equation: $\int_{\pi /2}^x \sqrt{3 - 2\sin^2 u} \,du + \int_0^y \cos t \,dt = 0.$ Applying the Leibniz rule for differentiation under the integral sign with respect to $x$: $\frac{d}{dx} \left( \int_{\pi /2}^x \sqrt{3 - 2\sin^2 u} \,du \right) + \frac{d}{dx} \left( \int_0^y \cos t \,dt \right) = 0.$ Using the Fundamental Theorem of Calculus: $\sqrt{3 - 2\sin^2 x} + \cos y \cdot \frac{dy}{dx} = 0.$ Rearranging the terms to solve for $\frac{dy}{dx}$: $\cos y \cdot \frac{dy}{dx} = -\sqrt{3 - 2\sin^2 x}.$ Therefore,$\frac{dy}{dx} = -\frac{\sqrt{3 - 2\sin^2 x}}{\cos y}.$ Thus,the correct option is $B$.

If $\int_{\pi /2}^x \sqrt{3 - 2\sin^2 u} \,du + \int_0^y \cos t \,dt = 0,$ then $\frac{dy}{dx} = $

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