Question

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

Answer 1

We are looking for a function whose derivative is $\sin 2x$
We recall,
$\frac{d}{dx}\cos 2x=-2\sin 2x$
$\Rightarrow \sin 2x=-\frac{1}{2}\frac{d}{dx}\left(\cos 2x\right)$
$\Rightarrow \sin 2x=\frac{d}{dx}(-\frac{1}{2}\cos 2x)$
Hence, the anti-derivative of $\sin 2x$ is $-\frac{1}{2}\cos 2x$