For what it's worth:
You can use the "guess and check" method as follows.
The derivative of $-\cos x$ is $\sin x$. So, perhaps the antiderivative of $5\sin(4x)$ is $ -5\cos(4x).$
Does this work? Let's check:
The derivative of our guess has to be $5\sin(4x)$; but, $ {d\over dx} \bigl(-5\cos (4x)\bigr)=5\sin(4x)\cdot 4= 5\cdot 4\sin(4x). $ Hmm, it's not quite right, we do not want that "4" there on the right hand side, that arose from the chain rule. But, if we introduced a multiplicative factor of $1\over4$ in our guess for the antiderivative, things would work out: $ {d\over dx} \bigl(-{5\over4}\cos (4x)\bigr)={5\over4} \sin(4x)\cdot 4= 5 \sin(4x). $
So, indeed, an antiderivative of $5\sin(4x)$ is $-{5\over4}\cos(4x)$.
More generally:
If $F(x)$ is an antiderivative of $f(x)$, then
$\ \ \ $1) $cF(x)$ is an antiderivative of $cf$ (since $(cf)'=c f'$)
$\ \ \ $2) For $c\ne0$, ${1\over c}F(cx)$ is an antiderivative of $f(cx)$ (by the chain rule).
Of course, you can use the "substitution method" for integrals for your problem.