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$$\sin 2x = 2\sin x \cos x \quad\text{ and }\quad -2\sin2x = -4\sin x \cos x.$$

However, when I integrate each with respect to $x$: $$\int (-2\sin 2x)dx=\cos 2x$$ and $$\int( -4\sin x\cos x)dx=2\cos^2x$$

Clearly $$\cos 2x ≠ 2\cos^2x.$$

Why is this the case when $$-2\sin 2x = -4\sin x\cos x?$$

Which one am I suppose to use when integrating?

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    Because cos(2x) and 2cos^2(x) are the same function, up to an additive constant (which should appear anyway in every primitive). – 2017-01-26
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    I edit you question for convenience of the readers. Is that what you mean? – 2017-01-26
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    As Did said, your two integrals **are** the same, you just forgot the "constant of integration". – 2017-01-26
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    @Did What happens if it's a definite integral (hence no constant)? Depending on the upper and lower limit, I might get a different answer depending on which function I choose to integrate over, right? – 2017-01-26
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    For limits $(a,b)$, one would get $$\left.\cos(2x)\right|_a^b=\cos(2b)-\cos(2a)$$ and/or $$\left.2\cos^2x\right|_a^b=2\cos^2b-2\cos^2a$$ **which coincide**. (Sorry but do you understand that $$\cos(2b)-\cos(2a)=2\cos^2b-2\cos^2a$$ for every $a$ and $b$?) – 2017-01-26

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This show the importance of adding a constant in the Indefinite Integral, notice that $$∫(−2\sin2x)dx=\cos2x+C_1$$ and $$\int( -4\sin x\cos x)dx=2\cos^2x+C_2$$ On the other hand you know the trigonometric identity $$\cos2x=2\cos^2x-1$$ Indeed for the above integrals just need an aproppiate constant to verify the identity!

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    What happens if it's a definite integral (hence no constant)? Depending on the upper and lower limit, I might get a different answer depending on which function I choose to integrate over, right? – 2017-01-26
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    If you have an definite integral that is correct! – 2017-01-26
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    Wow, so certain definite integrals give multiple answers; genuinely never knew that before! – 2017-01-26
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    Its obvious that integrate a diferents functions in an fix interval give multiples answers.. don't see any new in that. I hope you're not refering to make definite integral of the same function using an identity, i wasn't reply to that.... – 2017-01-26