Primitive of $\sin(4x)$ If $\sin(4x)=2\sin(2x)\cos(2x)$ then:
$$\int \sin(4x)=\int 2\sin(2x)\cos(2x)$$
$$\frac{-\cos(4x)}{4}=\frac{-\cos^2(2x)}{2}$$
But considering now $x=2$:
$$\frac{-\cos(4\cdot 2)}{4}\neq\frac{-\cos^2(2\cdot 2)}{2}$$
$$0.036\neq -0.214$$
What's the error that is behind this process ?
Thanks in advance.

Don't forget that an indefinite integral is only determined up to an additive constant!
You have
$$ \cos{4x}=2\cos^2{2x}-1, $$
so both expressions really do differ by a constant. If you evaluate them at both ends of a range (as in evaluating a definite integral), you will find this constant cancels out.
