# Different results in integrating both sides of $\sin{2x}=2\cos x\sin x$

I feel like there is something I am missing here. When integrating both sides of the trigonometric identity $\sin{2x}=2\cos x\sin x$ I get different results.

The left side of course results in $-\frac{1}{2}\cos{2x}+C$.

The right side I solve with u-substitution:

$u=\cos x$

$du=-\sin x dx$

$-2\int udu=-u^2+C=-\cos^2 x+C$

While writing this question I noticed another identity $\cos^2 x=\frac{1}{2}+\frac{1}{2}\cos 2x$. So apparently the $\frac{1}{2}$ falls out because of the $+C$ resulting from indefinite integration? This is still a little confusing to me.

You are correct to recall that $\cos^2x=\frac{1+\cos 2x }{2}$.

This is an indefinite integral. So, the constant term $\frac 12$ is not relevant. That is

$$\int \sin 2x \,dx=-\frac12\cos (2x)+C_1 \tag 1$$

and

\begin{align} \int \sin 2x \,dx&=-\cos^2 x+C_2\\\\ &=-\frac12\cos 2x+(-\frac12 +C_2)\\\\ &=-\frac12\cos 2x+C_3\tag 2 \end{align}

where we absorbed the constants $-\frac12+C_2$ into a new constant and called that new constant $C_3$. Inasmuch as the integration constant is arbitrary, $(1)$ and $(2)$ are equivalent statements.

• Thanks, it's clear now. I was thrown off by the cosines having a different power. – Rubenknex Sep 3 '15 at 20:04
• You're welcome. My pleasure. Although it can be confusing, you were on the right track. – Mark Viola Sep 3 '15 at 20:30

Integration process gives you results correct up to an arbitrary constant.

Both results are essentially the same.

$$- \frac{\cos 2 x}{2} + C_1 = - \frac { 2 \cos ^2 x -1}{2} + C_1 = - \cos ^2 x + C_2$$

That is why it is better to write $C_1, C_2$ for the integration constants.