alleryUsing inspection, an antiderivative of $\cos 2x$ is
A$2\sin 2x$
B$\dfrac{1}{2}\cos 2x$
C$-\dfrac{1}{2}\sin 2x$
D$\dfrac{1}{2}\sin 2x$
Answer & Solution
Correct answer: D. $\dfrac{1}{2}\sin 2x$
1. We seek $F$ with $F'(x)=\cos 2x$.
2. $\dfrac{d}{dx}(\sin 2x)=2\cos 2x$.
3. So $\cos 2x=\dfrac{d}{dx}\left(\dfrac{1}{2}\sin 2x\right)$, giving antiderivative $\dfrac{1}{2}\sin 2x$.
_Source: syllabus Mathematics Ch 7 "Integrals", p.292_
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