$$ \int_{\pi/4}^{3\pi/4} x \cdot \cot(x) \cdot \csc(x) \mathop{dx} $$
Some working
Evaluate the indefinite integral first
Let $u = \csc(x)$ , then $\frac{du}{- \csc(x) \cdot \cot(x)} = dx$, and $\csc^{-1} u = x$
This gives
\begin{equation*} \begin{aligned} \int x \cdot \frac{\cot(x) \cdot \csc(x) }{\cot(x) \cdot \csc(x)} \cdot (-1) \mathop{du} &= \int \csc^{-1} (u) (-1) \mathop{du} \\ &= - \int \csc^{-1} (u) \mathop{du} \end{aligned} \end{equation*}
Then using integration by parts
\begin{equation*} \begin{aligned} - \int \csc^{-1} (u) \mathop{du} &= - \left( u \cdot \csc^{-1}(u) - \int u \cdot \frac{1}{u \sqrt{u^2 - 1}} \right) \\ &= - u \cdot \csc^{-1}(u) + \int u \cdot \frac{1}{u \sqrt{u^2 - 1}} \\ &= - u \cdot \csc^{-1}(u) + \int \frac{1}{\sqrt{u^2 - 1}} \end{aligned} \end{equation*}
For the integral $\int \frac{1}{\sqrt{u^2 - 1}}$ let $k = u^2$, then $\frac{dk}{2u} = du$
\begin{equation*} \begin{aligned} \int \frac{1}{\sqrt{ u^2 - 1}} &= \int \frac{1}{\sqrt{k - 1}} \cdot \frac{1}{2 \sqrt{k}} \mathop{dk} \\ &= \frac{1}{2} \int \frac{1}{ \sqrt{ k} \sqrt{k - 1}} \mathop{dk} \\ &= \frac{1}{2} \sec^{-1} \left( \sqrt{ k} \right) \\ &= \frac{1}{2} \sec^{-1} \left( u \right) \end{aligned} \end{equation*}
. . .
I'm not sure what's going wrong - the rest of the working is messy, and I can't seem to find the error by checking or using CAS.
HINT:
Integrate by parts with $u=x$ and $v=-\csc(x)$. This yields
$$\int_{\pi/4}^{3\pi/4} x\cot(x)\csc(x)\,dx= \left.\left (-x\csc(x)\right)\right|_{\pi/4}^{3\pi/4}+\int_{\pi/4}^{3\pi/4}\csc(x)\,dx$$
Can you finish?