1
$\begingroup$

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?

  • 1
    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 $\sin ax$.2012-05-14

3 Answers 3

3

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.

2

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

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.

0

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

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

  • 0
    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!2013-03-25