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I don't understand... When I take a definite integral from a to b of 4cos(x)sin(x), u-substitution tells me that the answer should be 2sin^2(x) evaluated from b - it's evaluation at a. But my first reflex when seeing this integral was to use the identity sin(2x) = 2cos(x)sin(x) to get 4cos(x)sin(x) = 2sin(2x) and therefore integrate on that. Doing so gives me an evaluation with similar boundaries of -cos(2x), so I thought -cos(2x) = 2sin^2(x). But I know this to be wrong from the double angle identity of cos(2x) = 1-2sin^2(x) => -cos(2x) = 2sin^2(x)-1. Where did the 1 go?

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$-\cos2x$ and $2\sin^2x$ have the same derivative, so they differ by a constant. That constant is $1$.

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  • $\begingroup$ Oh of course, I feel incredibly dumb. I had thought about this but then told myself that the integral value wouldnt be the same, but since we are evaluation from one bound to the other the 1s end up cancelling out... Thank you! $\endgroup$ – Little Narwhal Apr 15 '20 at 14:59

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