I was proving this equation in class but I ran into a problem $$\int_0^\pi u^2dx \leq \int_0^\pi (u')^2dx$$ I have $$0 \leq \int_0^\pi (u' - u \cot(x))^2 dx = \int u'-2uu'\cot(x) + u^2\cot^2(x)dx$$ $$\int_0^\pi 2uu'\cot(X) - u^2\cot^2(x)dx \leq \int_0^\pi (u')^2dx$$ $$\int_0^\pi (u^2)'\cot(x) - u^2\cot^2(x)dx \leq \int_0^\pi (u')^2dx$$ Integrate by parts $$ u^2\cot(x)|_0^\pi + \int_0^\pi u^2\csc^2(x) - u^2 \cot^2(x) dx \leq \int_0^\pi(u')^2 dx$$
I know that $u^2\cot(x)|_0^\pi$ is an improper integral
My teacher suggested: $$^1 \lim_{x\rightarrow 0} u^2\cot(x) = ^2\lim_{x\rightarrow 0} \frac{u^2}{\sin(x)} = ^3 \lim_{x\rightarrow 0} \frac{2uu'}{\cos(x)} = 0$$
Now I understand how the equation would play out if the above happens to be 0 $$\int_0^\pi u^2(\csc^2(x) - \cot^2(x))dx \leq \int_0^\pi (u')^2 dx$$ $$\int_0^\pi u^2dx \leq \int_0^\pi (u')^2dx$$ done
Where I get in trouble is when I got from 1 to 2 in my teachers suggestion. (2 to 3 makes sense its is just l'hopital's rule.) Why are you allows to convert to $\sin(x)$?