Definitely Violent

Let $\displaystyle I = \frac {\displaystyle \int_0^\frac \pi 2 \frac x{\sin x}dx + \int_0^\frac \pi 2 \ln(1+\tan x)\ dx - \int_0^\frac \pi 4 \ln \left((1+\tan x)^2\right) dx - \int_0^\frac \pi 4 \ln(\tan x)\ dx}{\displaystyle \int_0^\frac \pi 2 \frac x{\tan x}dx + \int_0^\frac \pi 2 \ln(\sin x)\ dx + \int_\frac \pi 4^\frac \pi 2 \ln(\tan x)\ dx } = \frac {I_1+I_2-I_3-I_4}{I_5+I_6+I_7}$

$\begin{aligned} I_3 & = \int_0^\frac \pi 4 \ln \left((1+\tan x)^2\right) dx = 2 \int_0^\frac \pi 4 \ln (1+\tan x)\ dx \end{aligned}$

$\begin{aligned} I_2 & = \int_0^\frac \pi 2 \ln(1+\tan x)\ dx \\ & = {\color{#3D99F6}\int_\frac \pi 4^\frac \pi 2 \ln(1+\tan x)\ dx} + \int_0^\frac \pi 4 \ln (1+\tan x)\ dx & \small \color{#3D99F6} \text{Using identity: } \int_a^b f(x)\ dx = \int_a^b f(a+b-x)\ dx \\ & = {\color{#3D99F6}\int_\frac \pi 4^\frac \pi 2 \ln \left(1+\tan \left(\frac {3\pi}4 -x\right)\right)\ dx} + \frac 12 I_3 & \small \color{#3D99F6} \text{Note that }\tan (\pi - \theta) = - \tan \theta \\ & = {\color{#3D99F6}\int_\frac \pi 4^\frac \pi 2 \ln \left(1-\tan \left(x + \frac \pi 4\right)\right)\ dx} + \frac 12 I_3 & \small \color{#3D99F6} \text{Let }u = x - \frac \pi 4 \implies du = dx \\ & = \int_0^\frac \pi 4 \ln \left(1-\tan \left(u + \frac \pi 2\right)\right)\ du + \frac 12 I_3 & \small \color{#3D99F6} \text{Note that }\tan (\pi - \theta) = - \tan \theta \\ & = \int_0^\frac \pi 4 \ln \left(1+\tan \left(\frac \pi 2-u\right)\right)\ du + \frac 12 I_3 \\ & = \int_0^\frac \pi 4 \ln (1+\cot u)\ du + \frac 12 I_3 \\ & = \int_0^\frac \pi 4 \ln \left(\frac {\tan u + 1}{\tan u}\right)\ du + \frac 12 I_3 \\ & = \int_0^\frac \pi 4 \ln (1+\tan u)\ du - \int_0^\frac \pi 4 \ln (\tan u)\ du + \frac 12 I_3 \\ & = I_3 - I_4 \end{aligned}$

$\begin{aligned} I_4 & = \int_0^\frac \pi 4 \ln (\tan x) dx & \small \color{#3D99F6} \text{By integration by parts} \\ & = x \ln (\tan x) \bigg|_0^\frac \pi 4 - \int_0^\frac \pi 4 \frac {x \sec^2 x}{\tan x} dx \\ & = 0 - {\color{#3D99F6}\lim_{x \to 0} x\ln(\tan x)} - \int_0^\frac \pi 4 \frac x{\sin x \cos x} dx & \small \color{#3D99F6} \text{by L'Hôpital's rule} \\ & = {\color{#3D99F6}\lim_{x \to 0} \frac {x^2\sec^ x}{\tan x}} - \color{#D61F06} \frac 12 \int_0^\frac \pi 4 \frac x{\sin 2x} dx & \small \color{#D61F06} \text{Let }u = 2x \implies du = 2\ dx \\ & = \lim_{x \to 0} \frac {x^2}{\sin x \cos x} - \color{#D61F06} \frac 12 \int_0^\frac \pi 2 \frac u{\sin u} du \\ & = \lim_{x \to 0} \frac {x}{\cos x} - \frac 12 I_1 \\ & = 0 - \frac 12 I_1 \end{aligned}$

$I_1 = - 2I_4$

$\begin{aligned} I_6 & = \int_0^\frac \pi 2 \ln(\sin x)\ dx & \small \color{#3D99F6} \text{By integration by parts} \\ & = x \ln(\sin x) \bigg|_0^\frac \pi 2 - \int_0^\frac \pi 2 \frac {x\cos x}{\sin x} dx \\ & = 0 - \int_0^\frac \pi 2 \frac x{\tan x} dx \\ & = - I_5 \end{aligned}$

$\begin{aligned} I_7 & = \int_\frac \pi 4^\frac \pi 2 \ln(\tan x)\ dx \\ & = {\color{#3D99F6} \int_0^\frac \pi 2 \ln(\tan x)\ dx} - \int_0^\frac \pi 4 \ln(\tan x)\ dx & \small \color{#3D99F6} \text{Using identity: } \int_a^b f(x)\ dx = \int_a^b f(a+b-x)\ dx \\ & = {\color{#3D99F6} \frac 12 \int_0^\frac \pi 2 \left[\ln(\tan x) + \ln \left(\tan \left( \frac \pi 2 - x\right) \right)\right] dx} - I_4 \\ & = \frac 12 \int_0^\frac \pi 2 \left[\ln(\tan x) + \ln (\cot x)\right] dx - I_4 \\ & = 0 - I_4 \end{aligned}$

Therefore, we have

$\begin{aligned} I & = \frac {I_1+I_2-I_3-I_4}{I_5+I_6+I_7} = \frac {-2I_4+I_3-I_4-I_3-I_4}{I_5-I_5-I_4} = \frac {-4I_4}{-I_4} = \boxed{4} \end{aligned}$