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I know that $$ \int \sin(x)\,dx=-\cos(x)+C. $$ But I am wondering what will be the $\int \sin(ax)$? I mean what if $x$ is being multiplied by a constant?

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Try to substitute $u=ax$ –  Bernhard May 14 '12 at 10:45
    
Sorry I dont get the idea, what is u? –  Sean87 May 14 '12 at 10:46
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How do you know that $\int \sin x dx=-\cos x+C$? You just know, or this is because $(\cos x+C)'=-\sin x$? If the latter, then try to guess what you need to differentiate to get $\sin ax$. –  Artem May 14 '12 at 10:47
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3 Answers

up vote 3 down vote accepted

Try the substitution $u=ax$ in the integral

$$ \int \sin(ax) dx $$

As you have $dx=du/a$, then you get

$$ \frac{1}{a} \int \sin(u)du $$

This you can integrate, after which you can do the back-substitution.

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$$\int f(kx)=\frac{g(kx)}{k} +c$$

where $\int f(x)=g(x)$

the more geneal formula is,

$$\int p(qx)=\frac{r(qx)}{q'(x)}+c$$

where $\int p(x)=r(x)$

You can prove these by differentiating both sides.

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Assuming that you have not yet covered formal substitution methods (the way you wrote your question suggests this), here's how you could be expected to find it.

Since you know $\int \sin(x)\,dx=-\cos(x)+C,$ a reasonable guess is that $\int \sin(ax)\,dx$ might be equal to $-\cos(ax) + C.$ However, when we check this by differentiating $-\cos(ax) + C,$ we get $a\sin(ax)$ and not $\sin(ax).$ [Remember, when you have $\int f(x)\,dx = g(x) + C,$ the derivative of $g(x) + C$ will be equal to $f(x)$.] However, we're not all that far off -- when we checked by differentiating, we got something that is $a$ times bigger than it is supposed to be, but otherwise it was fine. This suggests that maybe using $a$ times smaller (i.e. $\frac{1}{a}$ times bigger) than our initial guess might work. So let's try $-\frac{1}{a} \cos(ax) + C$ instead. We check to see if this works by differentiating $-\frac{1}{a} \cos(ax) + C.$ This time we get $\sin(ax),$ which is what we wanted. Thus, we have found that

$$\int \sin(ax)\,dx \; =\; -\frac{1}{a}\cos(ax)+C$$

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I didn't realize until I had finished that this question was asked nearly a year ago and was only being edited now, which put it among the newly asked questions! –  Dave L. Renfro Mar 25 '13 at 21:49
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