Integration inequality of double angle identity.

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I am having trouble grasping why the integrals of $2$ sides of a double angle identity are not equal to each other. Given the following identity:

$$\sin(2x) = 2\sin(x)\cos(x)$$

$$\int \sin(2x)dx = -0.5\cos(2x)$$

$$\int 2\sin(x)\cos(x)dx = -\cos^2(x) \text{ (using u substitution)}$$

plug in $\pi$ for $x$ and it is clear that $-0.5\cos(2x) \ne -\cos^2(x)$

Why is this ?

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You're doing the freshman mistake of not including a $+C$ after your integrals. Any function that can be written in the form $-0.5 \cos(2x)+C$ can be written in the form $-\cos^2 (x)+C$, albeit with a different $C$: $$-{1\over2}\cos(2x) + C = -{1\over2}(2 \cos^2 x - 1) + C = -\cos^2x + {1\over2}+C = -\cos^2x + C'. $$