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(A) The anti-derivative of $\cos 3x$ is a function $F(x)$ such that $F'(x) = \cos 3x$. We know that the derivative of $\sin 3x$ with respect to $x$ is

Vedclass · Accepted Answer

$\frac{1}{3} \sin 3x$

Vedclass · Answer

(A) The anti-derivative of $\cos 3x$ is a function $F(x)$ such that $F'(x) = \cos 3x$. We know that the derivative of $\sin 3x$ with respect to $x$ is: $\frac{d}{dx}(\sin 3x) = 3 \cos 3x$ Dividing both sides by $3$,we get: $\frac{1}{3} \frac{d}{dx}(\sin 3x) = \cos 3x$ Using the linearity of the derivative,we can write: $\frac{d}{dx} \left( \frac{1}{3} \sin 3x \right) = \cos 3x$ Thus,the anti-derivative of $\cos 3x$ is $\frac{1}{3} \sin 3x$.

Find an anti-derivative (or integral) of the function $\cos 3x$ by the method of inspection.

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