[Math] Integrate $\int \sin (2x) \cos (2x)\,dx$

calculusindefinite-integralsintegrationtrigonometry

I have

$$\int \sin(2x) \cos (2x)\,dx = \frac12 \int \sin(4x)\,dx = -\frac18 \cos(4x),$$

but I also have

$$\int \sin(2x) \cos (2x)\,dx = \frac12 \int \sin 2x \cdot 2 \cos 2x \, dx = \frac14 \sin^2(2x).$$

Which one is correct, and why is the other method wrong?

Best Answer

They are differing by an integration constant, because of $\cos(2 y) = \cos^2(y) - \sin^2(y) = 1 - 2 \sin^2(y)$, and hence are the same as indefinite integration produces an anti-derivative up to a constant

Related Question