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I was given this problem \(\int sin(2x) dx\) and being the rebel I am refused to do u-substitution. Instead I did integration by parts. Here is my work and I think it is wrong but can not tell why.

\(\int sin(2x) dx=2\int sin(x)cos(x) dx\)

BY PARTS where f = sin(x) and g' = cos(x), f' = cos(x) and g = sin(x). The formula is \(\int fg' = fg-\int f'g\)

\(2\int sin(x)cos(x) dx = sin^2(x)-\int cos(x)sin(x)dx\)

\(3 \int sin(x)cos(x) dx = sin^2(x)\)

\(2\int sin(x)cos(x) dx = \frac{2sin^2(x)}{3}\)

\(\int sin(2x) = \frac{2sin^2(x)}{3}\)

 

What did I do wrong?

 Feb 3, 2020
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The first line of your integration, the one after the line BY PARTS .... , the RHS should be multiplied by 2.

You should also include constants of integration to tie the alternative results together via the trig identity for cos(2x).

 Feb 3, 2020

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