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?
$\int \sin(ax) \,\mathrm d x$
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1How 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 $\sin ax$. – 2012-05-14
3 Answers
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.
$\int f(kx)=\frac{g(kx)}{k} +c$
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.
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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0I 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! – 2013-03-25