Integral of Chebyshev polynomial expression $\int_{0}^{1} x^{-1/3}\cos(3\arccos(x))dx$

Question

I am trying to integrate

$$\int_{0}^{1} x^{-1/3}\cos(3\arccos(x))dx$$

but I am getting nowhere. The usual $$x=\cos(\theta)$$ substitution and other similar ones I have tried do not seem to work.

Answer 1

hint

For each real $ X,$ we have

$$\cos(3X) = 4\cos^3(X)-3\cos(X)$$

and for each $ Y\in [-1,1] $

$$\cos(\arccos(Y))=Y$$

The improper integral becomes

$$\int_0^1x^{-\frac 13}(4x^3-3x)dx=$$ $$4.\frac{3}{11}-3.\frac 35=\frac{-39}{55}$$