Dimensional analysis of $\int x \cos x dx$

I'm confused about using dimensional analysis, as in Street-Fighting Mathematics, of integrals like $$\int \! x \cos x \, \textrm{d}x = \textrm{something}$$

1. I start by expressing the right side as a function of one or more variables. It seems this should be a function of $x$ but it might additionally be a function of $\cos x$, so let's just say $x$ for now $$\int \! x \cos x \, \textrm{d}x = f(x)$$

2. I then assign dimensions to $x$ but this is confusing because the first $x$ is used as a scaling factor and the second $x$ is used as an angle (in radians or in degrees, I don't think it matters at this point). Let's just say $x$ is a length in both cases $$[x] = L$$

3. I then find the dimensions of the integral but this is also confusing because I'm multiplying a dimension with the cosine of the same dimension. Let's just assume $[\cos x] = L$ $$\left[\int \! x \cos x \, \textrm{d}x \right] = [x][\cos x][\textrm{d}x] = L^3$$

4. I finally make $f(x)$ with the same dimensions as the integral. Because the dimensions of $x$ are $L$, I have $$f(x) \sim L^3$$

If I compare this result with the actual antiderivative, I'm really confused because I'm adding different dimensions irrespective of the constant of integration $$[x\sin x + \cos x + C] = L^2 + L + ?$$

If I compare with the series expansion, then it seems to make sense when $n=1$ $$[x \sum_{n=0}^{\infty} \frac{(-1)^nx^{2n}}{(2n)!}] = L \times L^2 = L^3$$

This is all very confusing though so any clarifications would be much appreciated.

The argument of the cosine is necessarily dimensionless (otherwise the series expansion, which involves adding different powers of $x$, would not work). Therefore $x$ is dimensionless, and thus the integral is dimensionless.