Prove upper bound from gaussian quadrature

Question

Let $I = \int_{0}^1 ln(1+sin(x))dx$, and P be the polynomial with degree less or equal than 5 using the minimum square method to approximate the function $ln(1+sin(x))$ on [0,1]. Show that $$2\underset{x\in[0,1]}{sup}|ln(1+sin(x))-P(x)|$$ is an upper bound of the maximum absolute error of I using gaussian quadrature with 3 points

Answer 1

I've tried to several things like: By the mean value theorem of integrals, there exists an $\alpha \in [0,1]$ such that $ln(1+sin(\alpha)) = I $, so if a is the gaussian quadratue with 3 points, $|I-a| = |I-P(x) + P(x) - a| \leq |I-P(x)| + |P(x) - a|$, if I could show that $|P(x) - a| \leq |I-P(x)|$ I'd have the problem solved, but can't progress from here...

Also, I tried to rewrite $I = \displaystyle \sum_0^5 1/(n+1)a_n $, but couldn't progress from here as well (this is the first equation to approximate f(x) using the minimum square method).