https://ww3.math.msu.edu/images/webwork/132/Examples/4.7.4a-eg.pdf
I'm very confused how this integral trig formula works.
(Just using $ cos x $ in this case)
$$ \frac{1}{k}cos(kx) $$
How come if the equation is for example $ \frac{1}{2}cos(2x) $ would come be how it is? My book never uses this formula in the book. However I was given this in a PDF randomly on a problem on the online homework.
By the fundamental theorem of calculus
$\frac {d}{dx} \int f(x) dx = f(x)$
So, we can just take the derivative of $-\frac 12 \cos 2x + C$ and see what we get back. Which is exactly what is going on the the attached document.
But that is a little bit unsatisfying.
Nonetheless, from first principles you must differentiate $\sin x$ and $\cos x$ and show that $\frac {d}{dx} \sin x = \cos x$ and $\frac {d}{dx} \cos x = -\sin x.$ Which I have to assume you have already done in class and in your book and I am not going to repeat the derivations here.
Then when confronted with: $\int \sin 2x\ dx $ You are going to need to do a substitution to make it take on the form of a function whose anti-derivative you know.
$\int \sin 2x\ dx = \int \frac 12 \sin u \ du = -\frac 12 \int \sin u\ du = -\frac 12 \cos u = -\frac 12 \cos 2x$