After working $-2 \displaystyle\int (\cos^2\,x-\sin^2\,x)\,\sin\,2x\,dx$, I got $\frac{-\sin^2(2x)}{2}+c$. The answer I was given was $\frac{\cos^2(2x)}{2}+c$. Are these both legitimate solutions, depending on if sine or cosine is chosen for substitution?
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$E=\frac{-\sin^2(2x)}{2}+c=\frac{-1+\cos^2(2x)}{2}+c$ $E=\frac{\cos^2(2x)}{2}+c-\frac 1 2=\frac{\cos^2(2x)}{2}+k$
Where $ k=c-\frac 1 2$
These are the same answers.
user577215664
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$c' +\frac{1}{2} = c$
– fGDu94 Nov 28 '19 at 03:25