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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?

  • 2
    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.)2011-09-27
  • 6
    They are both wrong. The constant of integration has been omitted in each case.2011-09-27

3 Answers 3