Can't find my error in integrating $\sin(1-2x)\cos(1+2x)$

Question

My goal is to integrate $$I = \int \sin(1-2x)\cos(1+2x)\mathrm dx.$$

First, I let $u = -x$ and $\mathrm du=-\mathrm dx,$ and note that $$I = -\int \sin(1+2u) \cos(1-2u)\mathrm du.$$

Next, I try to integrate $I = \int \sin(1-2x)\cos(1+2x)\mathrm dx$ by parts using $u = \sin(1-2x)$ and $\mathrm dv=\cos(1+2x)\mathrm dx,$ which gives $\mathrm du = -2\cos(1-2x)$ and $v=\frac{\sin(1+2x)}{2}$. So, $$I = \frac{\sin(1-2x)\sin(1+2x)}{2} - \int\frac{\sin(1+2x)}{2} (-2\cos(1-2x)) \mathrm dx\\= \frac{\sin(1-2x)\sin(1+2x)}{2} + \int\sin(1+2x)\cos(1-2x)\mathrm dx.$$

But $\int\sin(1+2x)\cos(1-2x)\mathrm dx = -I,$ so $$I= \frac{\sin(1-2x)\sin(1+2x)}{2} - I,$$ and so $$I = \frac{\sin(1-2x)\sin(1+2x)}{4} + C.$$

However, symbolab and and wolframalpha give different answers that don't seem to just differ by a constant. I'm having a hard time finding my mistake.

Answer 1

"But $\int\sin(1+2x)\cos(1-2x)\mathrm dx = -I$"

NO. What we have is $$\int\sin(1+2u)\cos(1-2u)\mathrm du = -I$$ and it is NOT equal to the previous integral. Actually, we have $$\int\sin(1+2u)\cos(1-2u)\mathrm du =\int\sin(1-2x)\cos(1+2x)\mathrm dx$$

The variables $u$ and $x$ can be interchanged freely for definite integrals, as long as the limits of integration remain the same; i.e. $$\int_{x=a}^b\sin(1+2x)\cos(1-2x)\mathrm dx=\int_{u=a}^b\sin(1+2u)\cos(1-2u)\mathrm du.$$

Answer 2

In cases like this one I prefer to use the Euler relations $\sin(x) = (e^{ix}-e^{-ix})/2i$ and $\cos(x) = (e^{ix}+e^{-ix})/2$. They are simple but very efficient.

Applying these to the integrand yields: $$\sin(1+2x)\cos(1-2x)= (\sin(2)-\sin(4x))/2$$

Integration is now straightforward. The result is: $$I = x\sin(2)/2 +\cos(4x)/8 + C$$