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

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


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

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.

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


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.