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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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    **Hint.** Set $u=\frac{x}{2}$ and do a substitution.2012-07-24
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    I actually just tried that before I asked this and it turned out ugly! :[2012-07-24
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    Then you are doing things wrong. It doesn't "turn ugly". Post your work and we'll tell you where you are messing up.2012-07-24
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    And how is that "ugly"? The factor of $2$ is multiplying, so you can pull it out **exactly** the same way you pulled out the $3$. And surely you know what $\int\sin(u)\,du$ is!2012-07-24
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    this particular problem is what I am struggling with. I missed a lesson so I'm pretty much learning substitution and integration at the same time. The integration isn't hard, It's just that this problem confused me because of the trig function with the fraction inside2012-07-24
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    The problem, I think, is that you don't know what $\int \sin(u)\,du$ is. Do you know it?2012-07-24
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    isnt it -cos(u) + c?2012-07-24
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    Oh, I just realized the problem. You think that $u$ "transforms into" $\frac{u^2}{2}$ is this case, mainly because you are confusing it with a polynomial. If it were right, then $\int\sin(u)\,du = -\cos(u^2/2) + C.$2012-07-24
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    You are transforming integration in a mechanic thing; be careful...2012-07-24
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    so If I were to solve for C, the answer in my book reads y = 1 + 3sqrt3 - 6cos(x/2). Could you show me how to solve for C in this case?2012-07-24
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    let us [continue this discussion in chat](http://chat.stackexchange.com/rooms/4216/discussion-between-kudla69-and-ian-mateus)2012-07-24
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    @Kudla69: Yes, $\int \sin(u)\,du = -\cos(u)+C$. **No**, it is not $-\cos(u^2/2)+C$. And I don't know where your book got $1+3\sqrt{3}$; that would only be the case if you are looking for a **specific** antiderivative, not just the integral. Did you only give us **part** of the problem, then? Did you have some condition saying that a value of $y$ was equal to something specific, and never told us?2012-07-24
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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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    [Like the waitress said](http://people.oregonstate.edu/~scarbost/huh/huh.html#Old%20Joke%20-%20Version%201), "plus a constant!"2012-07-24
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    @ArturoMagidin Thanks for sharing this link! I had a good time.2012-07-24