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(A) The anti-derivative of $\sin 2x$ is a function $F(x)$ such that $F'(x) = \sin 2x$. We know that the derivative of $\cos 2x$ is: $\frac{d}{dx}(\cos

Vedclass · Accepted Answer

$-\frac{1}{2} \cos 2x$

Vedclass · Answer

(A) The anti-derivative of $\sin 2x$ is a function $F(x)$ such that $F'(x) = \sin 2x$. We know that the derivative of $\cos 2x$ is: $\frac{d}{dx}(\cos 2x) = -2 \sin 2x$ Dividing both sides by $-2$,we get: $-\frac{1}{2} \frac{d}{dx}(\cos 2x) = \sin 2x$ Using the linearity property of the derivative,we can write: $\frac{d}{dx}\left(-\frac{1}{2} \cos 2x\right) = \sin 2x$ Thus,the anti-derivative of $\sin 2x$ is $-\frac{1}{2} \cos 2x$.

Find an anti-derivative (or integral) of the function $\sin 2x$ by the method of inspection.

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