I am to solve the following initial value problem:
$$\frac {dy}{dx}=(4+y)\cos x + 4x(4+y),\ y(0)=0$$

I am aware that the general formula for an ODE/IVP is:
$$\frac {dy}{dx}+p(x)\cdot y=q(x),$$
and the integrating factor is:
$$I(x)=e^{\int p(x)\ dx}$$
However, do I expand the terms and solve like this:
$$\frac {dy}{dx}-\cos\ x\cdot y=4\cos x+4x(4+y)$$
so there are $p(x)$ and $y$ terms on the left and a $q(x)$ term on the right?

Integrating that is proving difficult and I think I have gone about arranging the equation wrong.

Thank you.

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**EDIT:**
This allows me to have:
$$e^{\int -\cos(x)\ dx}$$
which is:
$$e^{-\sin(x)}$$