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# Integration Problem.

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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