Log Trig Integral 6

$\begin{aligned} I & = \int_0^\frac \pi 2 \log(\tan x + \cot x) dx \\ & = \int_0^\frac \pi 2 \log \left(\frac 1{\sin x \cos x}\right) dx \\ & = \int_0^\frac \pi 2 \log \left(\frac 2{\sin 2x}\right) dx \\ & = \int_0^\frac \pi 2 \log 2 \ dx - \color{#3D99F6} \int_0^\frac \pi 2 \log (\sin 2x ) dx & \small \color{#3D99F6} \text{See note: } \int_0^\frac \pi 2 \log (\sin 2x ) dx = - \int_0^\frac \pi 2 \log 2 dx \\ & = 2 \int_0^\frac \pi 2 \log 2 \ dx = \pi \log 2 \end{aligned}$

$\implies a = \boxed{2}$

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Note:

$\begin{aligned} I_1 & = \int_0^\frac \pi 2 \log (\sin 2x ) dx & \small \color{#3D99F6} \text{Let } u = 2x \implies du = 2\ dx \\ & = \frac 12 \int_0^\pi \log (\sin u ) du & \small \color{#3D99F6} \text{Since } \sin u \text{ is symmetrical about }\frac \pi 2 \\ & = \int_0^\frac \pi 2 \log (\sin u ) du & \small \color{#3D99F6} \text{Replacing }u\text{ with }x \\ & = \int_0^\frac \pi 2 \log (\sin x ) dx \end{aligned}$

Also that:

$\begin{aligned} I_1 & = \int_0^\frac \pi 2 \log (\sin 2x ) dx \\ & = \int_0^\frac \pi 2 \log (2\sin x \cos x) dx \\ & = \int_0^\frac \pi 2 \log 2 \ dx + \int_0^\frac \pi 2 \log (\sin x ) dx + \color{#3D99F6} \int_0^\frac \pi 2 \log (\cos x ) dx & \small \color{#3D99F6} \text{By } \int_a^b f(x) \ dx = \int_a^b f(a+b-x) \ dx \\ & = \int_0^\frac \pi 2 \log 2 \ dx + \int_0^\frac \pi 2 \log (\sin x ) dx + \color{#3D99F6} \int_0^\frac \pi 2 \log (\sin x ) dx \\ & = \int_0^\frac \pi 2 \log 2 \ dx + 2I_1 \\ \implies I_1 & = - \int_0^\frac \pi 2 \log 2 \ dx \end{aligned}$