# Integrate $\int \sin (2x) \cos (2x) dx$

I have

$\int \sin(2x) \cos (2x) = 1/2 \int \sin(4x) dx = -\cos(4x)/8$

But I also have

$\int \sin(2x) \cos (2x) = 1/2 \int \sin 2x \cdot 2 \cos 2x \; dx = (1/4) \sin^2(2x)$

which is correct, and why is the other method wrong?

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They're both correct. Why assume otherwise? With trigonometric functions things that look different are often the same. For example, $\sin^2x$ is the same as $1-\cos^2 x$. (In this case, the two differ from each other by a constant, and that's what you expect of two antiderivatives of the same function.) – Michael Hardy Sep 27 '11 at 0:17
They are both wrong. The constant of integration has been omitted in each case. – André Nicolas Sep 27 '11 at 1:11

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

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To illustrate the phenomena in your calculation, I've chosen the number $\pi/8$ as the lower boundary.

\begin{align*} \int_{\frac{\pi}{8}}^{x} \sin(4t) \,\mathrm{d}t &= \int_{\frac{\pi}{8}}^{x} 2 \sin(2t) \cos(2t) \,\mathrm{d}t \\ &= \int_{\frac{\pi}{8}}^{x} \sin(2t) \,\mathrm{d}(\sin(2t)) \\ \left . -\frac{\cos(4t)}{4} \right\rvert_{\frac{\pi}{8}}^{x} &= \left . \frac{\sin^2(2t)}{2} \right\rvert_{\frac{\pi}{8}}^{x} \\ -\frac{\cos(4x)}{4} + \color{red}{\frac{\cos(4\cdot\frac{\pi}{8})}{4}} &= \frac{\sin^2(2x)}{2} - \color{red}{\frac{\sin^2(2\cdot\frac{\pi}{8})}{2}} \\ -\frac{\cos(4x)}{4} + \color{red}{0} &= \frac{\sin^2(2x)}{2} + \color{red}{\frac{1}{4}} \end{align*}

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