$\zeta(2) = \frac{\pi^2}{6}$

Rewrite $\begin{aligned} \int_0^{\Large\frac{\pi}{2}}\tan x\ln(\sin x)\,dx=\int_0^{\frac{\pi}{2}}\frac{\sin x}{\cos x}\ln(\sin x)\,dx. \end{aligned}$ Let $\,y=\cos x$ , then $\,dy=-\sin x\,dx$ . For $\,0<x<\frac{\pi}{2}$ , we have $\,0<y<1$ . Now, it turns out to be $\begin{aligned} \int_0^{\Large\frac{\pi}{2}}\frac{\sin x}{\cos x}\ln(\sin x)\,dx&=\int_0^{\Large\frac{\pi}{2}}\frac{\sin x}{\cos x}\ln(\sqrt{1-\cos^2 x})\,dx\\ &=-\frac{1}{2}\int_0^1\frac{\ln(1-y^2)}{y}\,dy.\tag1\\ \end{aligned}$ Next, use Maclaurin series for natural logarithm : $\ln(1-y^2)=-\sum_{n=1}^\infty \frac{y^{2n}}{n}.\tag2\\$ Substituting $\,(2)$ to $\,(1)$ yields $\begin{aligned} \frac{1}{2}\int_0^1\frac{\ln(1-y^2)}{y}\,dy&=-\frac{1}{2}\int_0^1\sum_{n=1}^\infty \frac{y^{2n}}{ny}\,dy\\ &=-\frac{1}{2}\sum_{n=1}^\infty\int_0^1 \frac{y^{2n-1}}{n}\,dy\\ &=-\frac{1}{4}\sum_{n=1}^\infty \left.\frac{y^{2n}}{n^2}\right|_{y=0}^1\\ &=-\frac{1}{4}\sum_{n=1}^\infty \frac{1}{n^2}. \end{aligned}$ The infinite series above is defined as Riemann zeta function $\,\zeta (2)=\frac{\pi^2}{6}$ . Thus, $\begin{aligned} \int_0^{\frac{\pi}{2}}\tan x\ln(\sin x)\,dx&=-\frac{1}{4}\sum_{n=1}^\infty \frac{1}{n^2}\\ &=-\frac{1}{4}\cdot \frac{\pi^2}{6}\\ &=-\frac{\pi^2}{24}\\ &\approx \boxed{-0.411234} \end{aligned}$

Here is another way to solve the integral without using Taylor series. Substituting $y=x^2$ , we obtain $\int_0^1\frac{\ln(1-x^2)}{x}\ dx=\frac12\int_0^1\frac{\ln(1-y)}{y}\ dy$ Using the fact that $\frac{\ln(1-y)}{y}=-\int_0^1\frac{1}{1-xy}\ dx$ then $\frac12\int_0^1\frac{\ln(1-y)}{y}\ dy=-\frac12\int_{y=0}^1\int_{x=0}^1\frac{1}{1-xy}\ dx\ dy.$ Using transformation variable by setting $(u,v)=\left(\frac{x+y}{2},\frac{x-y}{2}\right)$ so that $(x,y)=(u-v,u+v)$ and its Jacobian is $2$ . Therefore $-\frac12\int_{y=0}^1\int_{x=0}^1\frac{1}{1-xy}\ dx\ dy=-\iint_A\frac{du\ dv}{1-u^2+v^2},$ where $A$ is the square with vertices $(0,0),\left(\frac{1}{2},-\frac{1}{2}\right), (1,0),$ and $\left(\frac{1}{2},\frac{1}{2}\right)$ . Exploiting the symmetry of the square, we obtain $\begin{aligned} \iint_A\frac{du\ dv}{1-u^2+v^2}&=2\int_{u=0}^{\Large\frac12}\int_{v=0}^u\frac{dv\ du}{1-u^2+v^2}+2\int_{u=\Large\frac12}^1\int_{v=0}^{1-u}\frac{dv\ du}{1-u^2+v^2}\\ &=2\int_{u=0}^{\Large\frac12}\frac{1}{\sqrt{1-u^2}}\arctan\left(\frac{u}{\sqrt{1-u^2}}\right)\ du+2\int_{u=\Large\frac12}^1\frac{1}{\sqrt{1-u^2}}\arctan\left(\frac{1-u}{\sqrt{1-u^2}}\right)\ du. \end{aligned}$ Since $\arctan\left(\frac{u}{\sqrt{1-u^2}}\right)=\arcsin u$ , and if $\theta=\arctan\left(\frac{1-u}{\sqrt{1-u^2}}\right)$ then $\tan^2\theta=\frac{1-u}{1+u}$ and $\sec^2\theta=\frac{2}{1+u}$ . It follows that $u=2\cos^2\theta-1=\cos2\theta$ and $\theta=\frac12\arccos u=\frac\pi4-\frac12\arcsin u$ . Thus $\begin{aligned} \iint_A\frac{du\ dv}{1-u^2+v^2} &=2\int_{u=0}^{\Large\frac12}\frac{\arcsin u}{\sqrt{1-u^2}}\ du+2\int_{u=\Large\frac12}^1\frac{1}{\sqrt{1-u^2}}\left(\frac\pi4-\frac12\arcsin u\right)\ du\\ &=\bigg[(\arcsin u)^2\bigg]_{u=0}^{\Large\frac12}+\left[\frac\pi2\arcsin u-\frac12(\arcsin u)^2\right]_{u=\Large\frac12}^1\\ &=\frac{\pi^2}{36}+\frac{\pi^2}{4}-\frac{\pi^2}{8}-\frac{\pi^2}{12}+\frac{\pi^2}{72}\\ &=\frac{\pi^2}{12} \end{aligned}$ and the result follows.

$\Large\color{#3D99F6}{\text{\# }\mathbb{Q}.\mathbb{E}.\mathbb{D}.\text{\#}}$

Relevant wiki: Polylogarithm

$\begin{aligned} I & = \int_0^\frac \pi 2 \tan x \ln (\sin x) \ dx \\ & = \int_0^\frac \pi 2 \frac {\sin x}{\cos x} \ln (\sin x) \ dx & \small \color{#3D99F6} \text{Let }u = \sin x \implies du = \cos x dx \\ & = \int_0^1 \frac {u \ln u}{1-u^2} \ du \\ & = \frac 12 \int_0^1 \left({\color{#3D99F6}\frac {\ln u}{1-u}} - {\color{#D61F06}\frac {\ln u}{1+u}} \right) du & \small \color{#3D99F6} \text{By }\int_a^b f(x) \ dx = \int_a^b f(a+b-x) \ dx \\ & = \frac 12 {\color{#3D99F6} \int_0^1 \frac {\ln (1-u)}u du} - \frac 12 {\color{#D61F06}\left( \ln(1+u)\ln x \bigg|_0^1 - \int_0^1 \frac {\ln (1+u)}u du \right)} & \small \color{#D61F06} \text{By integration by parts} \\ & = - \frac {\color{#3D99F6}\text{Li}_2 (1)}2 - \frac 12 \left(0 - {\color{#D61F06}\int_0^{-1} \frac {\ln (1-t)}t dt} \right) & \small \color{#3D99F6} \text{Li}_2(z) \text{ is dilogarithm function.} \\ & = - \frac {\color{#3D99F6}\text{Li}_2 (1)}2 - \frac {\color{#D61F06}\text{Li}_2 (-1)}2 & \small \color{#3D99F6} \text{Let }t = -u \\ & = - \frac {\pi^2}{12} + \frac {\pi^2}{24} = - \frac {\pi^2}{24} \approx \boxed{- 0.411} \end{aligned}$