Long Integral equals (π/2)^2−[2ln(2)/]π^2+ln^2(2)/(2π)

[Tony1]

How to show that,

$$\int_{0}^{1}t^2\sec(t\pi)[1+\sin(t\pi)-\sin^2(t\pi)]\ln[\sin(t\pi)]\mathrm dt=\left({\pi\over 2}\right)^2-{\color{blue}{2\ln(2)\over \pi^2}}+\color{red}{{\ln^2(2)\over 2\pi}}$$

 

[jvicens]

Hello,

To show this integral, we can start by using the trigonometric identity $\sin^2(t\pi) = \frac{1}{2}(1-\cos(2t\pi))$. Then, we can rewrite the integral as:

$$\int_{0}^{1}t^2\sec(t\pi)\left[1+\sin(t\pi)-\frac{1}{2}(1-\cos(2t\pi))\right]\ln[\sin(t\pi)]\mathrm dt$$

Next, we can use the substitution $u = t\pi$ and $du = \pi\mathrm dt$ to simplify the integral:

$$\frac{1}{\pi}\int_{0}^{\pi}u^2\sec(u)\left[1+\sin(u)-\frac{1}{2}(1-\cos(2u))\right]\ln[\sin(u)]\mathrm du$$

Using the power rule and the product rule for integration, we can expand the integral to:

$$\frac{1}{\pi}\int_{0}^{\pi}\left(u^2+u^2\sin(u)-\frac{1}{2}(u^2-u^2\cos(2u))\right)\sec(u)\ln[\sin(u)]\mathrm du$$

Simplifying further, we get:

$$\frac{1}{\pi}\int_{0}^{\pi}\left(u^2+u^2\sin(u)-\frac{1}{2}u^2+\frac{1}{2}u^2\cos(2u)\right)\sec(u)\ln[\sin(u)]\mathrm du$$

Using the trigonometric identity $\cos(2u) = 1-2\sin^2(u)$, we can rewrite the integral as:

$$\frac{1}{\pi}\int_{0}^{\pi}\left(\frac{1}{2}u^2+u^2\sin(u)-\frac{1}{2}u^2+u^2\sin^2(u)\right)\sec(u)\ln[\sin(u)]\mathrm du$$

Simplifying further, we get:

$$\frac{1}{\pi}\int_{0}^{\pi}\left(u^2+u^2\sin(u)-\frac{