Solve Laplace's equation in polar coordinates $$ \frac {1}{r} \frac {\partial u} {\partial r} + \frac {\partial^2 u} {\partial r^2} + \frac {1} {r^2} \frac {\partial^2 u} {\partial \theta^2} = 0$$ on the disk $$ {{(r, \theta) | 0 \leq r \leq R , 0 \leq \theta \leq 2}} $$ subject to the boundary condition $ u (R, \theta) = Tsin^2 (\theta) $
I got $ u (r, \theta) = \sum_{n=0}^{\infty}r^n [a_ncos (n\theta)+b_nsin (n\theta)] $ for $ n \in \mathbb{N} $
And solvin for the condition using
$$ a_n= 1/\pi \int_{0}^{2\pi} Tsin^2 (\theta)cos (n\theta) d\theta $$and
$$ b_n= 1/\pi \int_{0}^{2\pi} Tsin^2 (\theta)sin(n\theta) d\theta $$
I get $$ a_n = \frac{2Tsin (2\pi n)}{4n \pi-n^3 \pi} $$ and $$ b_n = \frac{4Tsin^2(\pi n)}{4n \pi-n^3 \pi} $$
But $ sin (n\pi)=0 $ for $n \in \mathbb{N} $ so i would get $ u (r, \theta) = 0 $
What should i consider for solving it?
You have to work a little harder to write down the exact values of $a_n$ and $b_n$.
$\int_0^{2\pi} sin ^{2}\theta \cos (n\theta) d\theta=\frac 1 2\int_0^{2\pi} (1-\cos (2\theta) \cos (n\theta) d\theta$ using the identity $cos (2\theta) \cos (n\theta)=\frac 1 2 (\cos (n+2) \theta -\cos (n-2) \theta$ this becomes $\pi \delta_{0,n} -\frac {\pi} 2 \delta_{2,n} -\frac {\pi} 2 \delta_{-2,n}$ wheer $\delta_{i.j}=1$ if $i=j$ and $0$ if $i \neq j$.
I will let you handle $b_n$ by a similar method. .