$\int \cos^2 (x\cdot \text{constant} )\, dx$ [Do I need to consider the constant?) I'm integrating $\cos^2 \left(\frac{\pi y}B\right) \,dy$, where $\frac{\pi}B$ is a constant.
Can I use this rule or does the face that I have a constant together with the X change the way of integration?
From the integral rule book form Rottmann
$$\int \cos^2 x \, dx = \frac12 \sin x \cos x+\frac{x}2+C$$
 A: It doesn't inherently change the overall idea. What you're basically trying to do is
$$\int \cos^2(Ax)dx$$
where $A = \pi/B$ is a constant. Then it's sort of like the reverse chain rule. Since you know
$$\int \cos^2(x)dx = \frac 1 2 \sin(x)\cos(x) + \frac x 2 + C$$
Just let $x \mapsto Ax$ but also be sure to divide by $A$ to account for the antidifferentiation. Then we have
$$\int \cos^2(Ax)dx = \frac 1 {2A} \sin(Ax)\cos(Ax) + \frac{Ax}{2A} + \frac{C}{A} = \frac 1 {2A} \sin(Ax)\cos(Ax) + \frac{x}{2} +C $$
(Note: slight abuse of notation there, since $C/A \ne C$. This is moreso just acknowledging that the constant of integration is arbitrary without initial conditions, so we can just replace the ratio of two constants, $C/A$, by another constant, and name it $C$ to appeal to our intuition on such matters.)

This overarching idea can be verified by making the $u$-substitution $u=Ax \implies du = Adx$, giving
$$\int \cos^2(Ax)dx = \int \cos^2(u) \frac{du}{A} = \frac 1 A \int \cos^2(u)du$$
A: To compute $\int \cos^2\left( cy\right)\, dy$, perform a substitution, $t=cy$, then $\frac{dt}{dy}=c$.
$$\int \cos^2\left( cy\right)\, dy = \frac1c\int \cos^2(t)\, dt$$
Now, you can use the formula and replace $t$ with $cy$ at the end.
