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In Griffiths electrodynamics in an example it evaluates the potential inside a hollow sphere of radius R having potential $V_0(\theta)$ potential inside is the form of this $$v(r,\theta) = \sum A_lr^lp_l(\cos\theta)$$ After solving the Laplace in spherical coordinates and applying orthogonality to find the constant $A_l$ we arrive at following integral

$$A_l = \frac{2l+1}{2R^l}\int_{0}^{\pi}V_0 P_l(\cos\theta)\sin\theta d\theta $$

(here $P_l(\cos\theta)$ is Legendre polynomial)

Now the book doesn't solve the above equation it just says that its very hard to find this integral so we apply a trick called "by eyeball" and assign $V_0 = k\sin^2(\frac{\theta}{2})$. then putting the value of $V_0(\theta)$ in this expression $$v(R,\theta) = \sum A_lr^lp_l(\cos\theta) $$

it evaluates $A_0$ and $A_1$ Now it solves the same question but this time we have to find potential outside the sphere then to find value of constant $B_l$ it just end the solution here

$$B_l = \frac{2l+1}{2} \int_{0}^{\pi} V_0(\theta)p_l(\cos\theta)\sin\theta d\theta$$

it just stops here it doesn't resort to that eyeball method. Why is it wrong to solve this integral by the "eyeball method" ?

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