(adsbygoogle = window.adsbygoogle || []).push({}); 1. This is part of a larger calculation. I do it one way and get the right answer and do it a seemingly equivalent way and get the wrong answer. The question is just about reconciling the two integrals below.

2.

[tex]I_1 = \int \frac{sin(2x)}{2} dx = -\frac{cos(2x)}{4} [/tex]

[tex] \frac{sin(2x)}{2} = sin(x)cos(x) [/tex]

[tex] I_2 = \int sin(x)cos(x) dx = - \frac{cos^2(x)}{2} [/tex]

For [tex] x= 0 [/tex], [tex] I_1 = -1/4 \neq I_2 = -1/2 [/tex].

How can this be?

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# Homework Help: Two (equivalent?) integrals

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