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$\int 3\sin\left(\frac{x}{2}\right)dx$

can't figure this one out! I'm not sure if I'm supposed to substitute or not?

Here's where I'm at...

$3\int\sin\left(\frac{x}{2}\right)dx$ $u = \frac{x}{2}$ $du = \frac{1}{2}dx$ $dx = 2du$ $3\int\sin\left(u\right)2du$

and then...

$-6cos\left(\frac{u^2}{2}\right)$ thats not right is it..?

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    @Kudla69: To verify that $\int\sin(u)\,du = -\cos(u)+C$, just take derivatives. If you were to take the derivative of $-\cos(u^2/2)+C$, you would get $\sin(u^2/2)(2u/2) = u\sin(u^2/2),$ by the Chain Rule. Definitely **not** $\sin(u)$. If you are having trouble with $\int\sin(u)\,du$, then your problems began waaaaay before you "missed a lesson". That kind of integral is pretty basic, and your error is pretty bad.2012-07-24

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Let $x=2t\implies dx=2dt$. Hence your problem becomes $2\int 3\sin(t)dt=-6\cos(t)+c=-6\cos(x/2)+c$ where $c$ is a constant.

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    @ArturoMagidin Thanks for sharing this link! I had a good time.2012-07-24