@utkFresh $$\int \sin(ax) dx $$
Let u=ax
So du=a dx
du/a=dx
So we have
$$\int \sin(u) \frac{du}{a} $$
\[\frac{1}{a} \int \sin(u) du \ \frac{1}{a} (-cos(u))+C \ frac{-1}{a}cos(ax)+C

Question

@utkFresh $$\int \sin(ax) dx $$
Let u=ax
So du=a dx
du/a=dx
So we have 
$$\int \sin(u) \frac{du}{a} $$
\[\frac{1}{a} \int \sin(u) du \ \frac{1}{a} (-cos(u))+C \ frac{-1}{a}cos(ax)+C

anonymous · Answer

$$\frac{1}{a} \int \sin(u) du \ \frac{1}{a} (-\cos(u))+C \ \frac{-1}{a}\cos(ax)+C $$

myininaya · Answer

$$\frac{1}{a} \int \sin(u) du \ \frac{1}{a} (-cos(u))+C \\ \frac{-1}{a}\cos(ax)+C 
  $$

anonymous · Answer

no charge

myininaya · Answer

no  I fixed it !

myininaya · Answer

I want a refund.

thesmartone · Answer

??? I am so confused.

thesmartone · Answer

Maybe its because I don't know Calculus...

myininaya · Answer

I just realized someone messaged me 26 minutes ago and they wanted to know if there is a formula.
a does not equal 0

thesmartone · Answer

oh yeah because then it would be undefined...