Integration of $\int_0^\frac{\pi}{4} (\sin^6 2x+\cos^6 2x) \cdot\ln (1+\tan x) dx$

Question

I've found an integration problem from Molodova Matholympiad 2008. The problem is as follows.

Find the Integration of $$\int\limits_0^\frac{\pi}{4} (\sin^6 2x+\cos^6 2x) \cdot \ln (1+\tan x)\ \mathrm dx$$

My approach is as follows.

Can anyone help me to find the full solution?

Answer 1

Let $f(x) = g(x)h(x)$ where $$g(x) = \sin^6 2x + \cos^6 2x, \quad h(x) = \log(1 + \tan x).$$ Then $$g(\pi/4 - x) = g(x), \\ h(\pi/4 - x) = \log\left(1 + \tan (\tfrac{\pi}{4} - x)\right) = \log \left( 1 + \frac{1 - \tan x}{1 + \tan x} \right) = \log \frac{2}{1 + \tan x} = \log 2 - h(x).$$ Consequently, $$f(\pi/4 - x) + f(x) = g(x) \log 2,$$ and we have $$\int_{x=0}^{\pi/4} f(x) \, dx = \frac{1}{2} \int_{x=0}^{\pi/4} f(x) + f(\pi/4-x) \, dx = \frac{\log 2}{2} \int_{x=0}^{\pi/4} g(x) \, dx.$$ Since $$\begin{align} a^6 + b^6 &= (a^2 + b^2)(a^4 - a^2 b^2 + b^4) \\ &= (a^2 + b^2)(a^4 + 2a^2 b^2 + b^4 - 3a^2 b^2) \\ &= (a^2 + b^2)\left((a^2 + b^2)^2 - 3a^2 b^2\right) \end{align}$$ it follows that $$\begin{align} g(x) &= 1 - 3 \sin^2 2x \cos^2 2x \\ &= 1 - \frac{3}{4} \sin^2 4x \\ &= 1 - \frac{3}{8} (1 - \cos 8x) \\ &= \frac{1}{8}(5 + 3 \cos 8x). \end{align}$$

Putting everything together, $$\int_{x=0}^{\pi/4} f(x) \, dx = \frac{\log 2}{16} \int_{x=0}^{\pi/4} 5 + 3 \cos 8x \, dx = \frac{5\pi \log 2}{64}.$$

Answer 2

This was done for the upper bound equal to $\frac \pi 2$ as written in the initial post.

Tedious but doable using Feynman's trick.

Consider $$I(a)=\int_0^{\frac \pi 2} \left(\sin ^6(2 x)+\cos ^6(2 x)\right) \log (1+a \tan (x))\,dx$$ $$I'(a)=\int_0^{\frac \pi 2} \frac{\tan (x) \left(\sin ^6(2 x)+\cos ^6(2 x)\right)}{1+a \tan (x)}\,dx$$ Now, using $x=\tan^{-1}(t)$ $$I'(a)=\int_0^{\infty} \frac{t^9-8 t^7+30 t^5-8 t^3+t}{\left(1+t^2\right)^5\, (1+a t)}\,dt$$ Using partial fraction decomposition, only two types of integrals remain $$J_n=\int_0^{\infty} \frac {dt}{(1+t^2)^n}=\frac{\sqrt{\pi } \,\,\Gamma \left(n-\frac{1}{2}\right)}{2 \Gamma (n)}$$ $$K_n=\int_0^{\infty} \frac {t}{(1+t^2)^n}\,dt=\frac{1}{2 (n-1)}$$ The final result is $$16 \left(a^2+1\right)^5\,I'(a)=16 \left(a^8-10 a^6+10 a^2-1\right)+$$ $$\pi a \left(5 a^8-4 a^6+198 a^4-148 a^2+29\right)-$$ $$16 \left(a^8-8 a^6+30 a^4-8 a^2+1\right) \log (a)$$

Again, a bunch of partial fraction decomposition and integration between $0$ and $1$ lead to $$\int_0^1 \frac{a^8-10 a^6+10 a^2-1}{\left(a^2+1\right)^5}\,dt=0$$ $$\int_0^1 \frac{\pi a \left(5 a^8-4 a^6+198 a^4-148 a^2+29\right)}{16 \left(a^2+1\right)^5}\,dt=\frac{5}{32} \pi \log (2)$$ $$\int_0^1 \frac{a^8-8 a^6+30 a^4-8 a^2+1 }{\left(a^2+1\right)^5} \log (a)\,dt=-\frac{5 C}{8}-\frac{1}{16}$$

$$\color{blue}{\int_0^{\frac \pi 2} \left(\sin ^6(2 x)+\cos ^6(2 x)\right) \log (1+ \tan (x))\,dx=\frac{5 }{8}C+\frac{1}{16}+\frac{5}{32} \pi \log (2)}$$

Edit

Changing the bound $$I'(a)=\int_0^{1} \frac{t^9-8 t^7+30 t^5-8 t^3+t}{\left(1+t^2\right)^5\, (1+a t)}\,dt$$ the same process does repeat and $$\color{blue}{\int_0^{\frac \pi 2} \left(\sin ^6(2 x)+\cos ^6(2 x)\right) \log (1+ \tan (x))\,dx=\frac{5}{64} \pi \log (2)}$$