If y = sqrt{cos x^2 + sqrt{cos x^2 + sqrt{cos | vedclass

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Question

(C) Given $y = \sqrt{\cos x^2 + y}$.Squaring both sides, we get $y^2 = \cos x^2 + y$.Differentiating both sides with respect to $x$:$2y \frac{dy}{dx}

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$\cos x^2 + c$

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(C) Given $y = \sqrt{\cos x^2 + y}$.Squaring both sides, we get $y^2 = \cos x^2 + y$.Differentiating both sides with respect to $x$:$2y \frac{dy}{dx} = -\sin x^2 \cdot (2x) + \frac{dy}{dx}$.Rearranging the terms:$(2y - 1) \frac{dy}{dx} = -2x \sin x^2$.Thus, $\frac{dy}{dx} = \frac{-2x \sin x^2}{2y - 1}$.Comparing with $\frac{dy}{dx} = \frac{f(x)}{2y - 1}$, we get $f(x) = -2x \sin x^2$.Now, $\int f(x) dx = \int -2x \sin x^2 dx$.Let $u = x^2$, then $du = 2x dx$.Substituting these, we get $\int -\sin u du = \cos u + c$.Substituting back $u = x^2$, we get $\cos x^2 + c$.

If $y = \sqrt{\cos x^2 + \sqrt{\cos x^2 + \sqrt{\cos x^2 + \dots \infty}}}$ and $\frac{dy}{dx} = \frac{f(x)}{2y - 1}$, then $\int f(x) dx = \dots$

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