$\int \sin(ax) \,\mathrm d x$ 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?
 A: 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.
A: $$\int f(kx)=\frac{g(kx)}{k} +c$$
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.
A: 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
$$\int \sin(ax)\,dx \; =\; -\frac{1}{a}\cos(ax)+C$$
