If int f(x) cos x , dx = frac{1}{2} [f(x)]^2 | vedclass

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(A) Given the equation: $\int f(x) \cos x \, dx = \frac{1}{2} [f(x)]^2 + C$. Differentiating both sides with respect to $x$ using the Fundamental Theo

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(A) Given the equation: $\int f(x) \cos x \, dx = \frac{1}{2} [f(x)]^2 + C$. Differentiating both sides with respect to $x$ using the Fundamental Theorem of Calculus, we get: $f(x) \cos x = \frac{d}{dx} \left( \frac{1}{2} [f(x)]^2 + C ight)$. Applying the chain rule: $f(x) \cos x = \frac{1}{2} \cdot 2 f(x) \cdot f'(x)$. Simplifying the expression: $f(x) \cos x = f(x) \cdot f'(x)$. For $f(x) 
eq 0$, we have $f'(x) = \cos x$. Substituting $x = 0$: $f'(0) = \cos(0) = 1$.

If $\int f(x) \cos x \, dx = \frac{1}{2} [f(x)]^2 + C$ and $f(0) = 0$, then $f'(0) = $

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