f^{prime}(x) = 3 sin x - 4 sin^3 x and f(0) = | vedclass

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(A) Given $f^{\prime}(x) = 3 \sin x - 4 \sin^3 x$. Using the trigonometric identity $\sin(3x) = 3 \sin x - 4 \sin^3 x$,we can rewrite the derivative a

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$\frac{2}{3}$

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(A) Given $f^{\prime}(x) = 3 \sin x - 4 \sin^3 x$. Using the trigonometric identity $\sin(3x) = 3 \sin x - 4 \sin^3 x$,we can rewrite the derivative as: $f^{\prime}(x) = \sin(3x)$. To find $f(x)$,we integrate $f^{\prime}(x)$ with respect to $x$: $f(x) = \int \sin(3x) \, dx = -\frac{\cos(3x)}{3} + c$. Given the initial condition $f(0) = \frac{1}{3}$,we substitute $x = 0$ into the equation: $f(0) = -\frac{\cos(0)}{3} + c = \frac{1}{3}$. Since $\cos(0) = 1$,we have: $-\frac{1}{3} + c = \frac{1}{3}$. Solving for $c$: $c = \frac{1}{3} + \frac{1}{3} = \frac{2}{3}$.

$f^{\prime}(x) = 3 \sin x - 4 \sin^3 x$ and $f(0) = \frac{1}{3}$,then $f(x) = c + \dots$ where $c$ is the constant of integration. Find the value of $c$.

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