Differential EquationsSuperchallengeBoundary Values & Fourier Series ShareBoundary Values & Fourier Series · Superchallenge problemShow that ln (2sinx2)=−∑n≥1cosnxn\ln\!\left(2\sin\dfrac x2\right) = -\displaystyle\sum_{n\ge1}\dfrac{\cos nx}{n}ln(2sin2x)=−n≥1∑ncosnx for 0<x<2π0 < x < 2\pi0<x<2π, and use Parseval's identity to evaluate ∫0πln2 (2sinx2)dx\displaystyle\int_0^{\pi}\ln^2\!\left(2\sin\dfrac x2\right)dx∫0πln2(2sin2x)dx.Check Show answer Walk through it step by step Practice more like this Review this concept