We have $$\frac{d}{d\phi}(sin\phi \frac{dP}{d\phi}) + \lambda sin\phi P =0 $$ By changing the variable in the equation to x = cos $\phi$, derive the self adjoint form of the Legendre equation: $$\frac{d}{dx}((1-x^2)\frac{dP}{dx}) + \lambda P$$
I know that in order to change the variables, $sin(\phi)=\sqrt{1-cos^2(\phi)}=\sqrt{1-x^2}$
So we have $$\frac{d}{dx}(\sqrt{1-x^2} \frac{dP}{dx}) + \lambda \sqrt{1-x^2} P =0 $$ $$\frac{1}{\sqrt{1-x^2}}\frac{d}{dx}(\sqrt{1-x^2} \frac{dP}{dx}) + \lambda P =0 $$ And I don't know where to go from here. Any help will be appreciated!
Since you have that $x=\cos \phi$ , then : $$\text { 1) }\frac {dx}{d\phi}=-\sin \phi$$ $$\frac {dx}{d\phi}=-\sqrt {1-\cos^2 \phi}=-\sqrt {1-x^2}$$ And also apply chain rule : $$\text { 2) }\frac {dP}{d\phi}=\frac {dP}{dx}\frac {dx}{d\phi} \text{ ,And , }\frac {d}{d\phi}=\frac {dx}{d\phi}\frac {d}{dx}$$ I think you can take it from there.