Integral of product = π/9[12ln(2)−1]

[Tony1]

How may one show that,

$$\int_{0}^{4\pi}\sin\left({x\over 2}\right)\sin\left({x\over 4}\right)\ln^2\left[\sin\left({x\over 8}\right)\right]\mathrm dx={\pi\over 9}[12\ln (2)-1]$$

 

[jvicens]

To show that this integral is equal to ${\pi\over 9}[12\ln (2)-1]$, we can use the properties of definite integrals and trigonometric identities.

First, we can use the property of definite integrals that states that $\int_{a}^{b}f(x)\mathrm dx=-\int_{b}^{a}f(x)\mathrm dx$. This allows us to rearrange the limits of integration to make the integral easier to evaluate.

Next, we can use the identity $\sin(a)\sin(b)={1\over 2}[\cos(a-b)-\cos(a+b)]$ to simplify the integrand. This will give us:

$$\int_{0}^{4\pi}\sin\left({x\over 2}\right)\sin\left({x\over 4}\right)\ln^2\left[\sin\left({x\over 8}\right)\right]\mathrm dx={1\over 2}\int_{0}^{4\pi}\cos\left({x\over 4}\right)\ln^2\left[\sin\left({x\over 8}\right)\right]\mathrm dx-{1\over 2}\int_{0}^{4\pi}\cos\left({3x\over 4}\right)\ln^2\left[\sin\left({x\over 8}\right)\right]\mathrm dx$$

Using the property of definite integrals again, we can combine these two integrals to get:

$$\int_{0}^{4\pi}\sin\left({x\over 2}\right)\sin\left({x\over 4}\right)\ln^2\left[\sin\left({x\over 8}\right)\right]\mathrm dx={1\over 2}\int_{0}^{4\pi}\cos\left({x\over 4}\right)\ln^2\left[\sin\left({x\over 8}\right)\right]\mathrm dx-{1\over 2}\int_{0}^{4\pi}\cos\left({3x\over 4}\right)\ln^2\left[\sin\left({x\over 8}\right)\right]\mathrm dx={1\over 2}\int_{0}^{4\pi}\cos\left