I'm reading "Principles of Ideal Fluid Aerodynamics" by Karamcheti. On page 285 he starts to construct the solution of
$$\tag{1}\frac{\partial}{\partial r}\left( r^2sin(\theta)\frac{\partial\phi}{\partial r}\right)+\frac{\partial}{\partial\theta}\left( sin(\theta)\frac{\partial\phi}{\partial\theta}\right)=0$$
where $\phi=\phi(r,\theta,t)$ using separation of variables. He assumes that $\phi(r,\theta)=R(r)\Theta(\theta)$. On the next page he separates this into two equations:
$$\tag{2}\frac{d}{dr}\left( r^2\frac{dR}{dr}\right)-kR=0$$
$$\tag{3}\frac{d}{d\theta}\left( sin(\theta)\frac{d\Theta}{d\theta}\right)+k\Theta\sin(\theta)=0$$
where k is a separation constant. He then sets $k=n(n+1)$ and $cos(\theta)=\mu$. Using these relationships, he claims (3) can be written as
$$\tag{4}\frac{d}{d\mu}\left[ (1-\mu^2)\frac{d\Theta}{d\mu}\right]+n(n+1)\Theta=0$$
Can anyone please help me understand how to go from equation (3) to (4)?
I think I answered my own question while writing it. Using the relationship $\mu=cos(\theta)$ we can write
$$d\theta=\tag{5}\frac{d\mu}{sin(\theta)}$$
which means (3) can be written as
$$\tag{6}sin(\theta)\frac{d}{d\mu}\left( sin^2(\theta)\frac{d\Theta}{d\mu}\right)+k\Theta\sin(\theta)=0$$
Using the relation
$$sin^2(\theta)=1-cos^2(\theta)=1-\mu^2$$
and dividing by $sin(\theta)$, we get (4).