Step 1: Simplify the denominator.
Using $1+\cos 2x=2\cos^2 x$, the integral becomes $I=\displaystyle\int_{\pi/4}^{3\pi/4}\dfrac{x\sin x}{2\cos^2 x}\,dx=\dfrac12\displaystyle\int_{\pi/4}^{3\pi/4} x\tan x\sec x\,dx$.
Step 2: Use the reflection property.
With limits $a=\dfrac{\pi}{4}$, $b=\dfrac{3\pi}{4}$, replace $x$ by $a+b-x=\pi-x$. Then $\sin(\pi-x)=\sin x$ and $\cos(\pi-x)=-\cos x$, so $\dfrac{\sin x}{\cos^2 x}$ keeps its form while the factor $x$ becomes $\pi-x$.
Step 3: Add the two versions.
Writing $I$ both ways and adding, $2I=\dfrac12\displaystyle\int_{\pi/4}^{3\pi/4}\pi\cdot\dfrac{\sin x}{\cos^2 x}\,dx$ after the $x$ terms combine, but carrying the sign from $\cos(\pi-x)$ flips it, giving $2I=-\dfrac{\pi}{2}\displaystyle\int_{\pi/4}^{3\pi/4}\sec x\tan x\,dx$.
Step 4: Integrate the standard piece.
$\displaystyle\int\sec x\tan x\,dx=\sec x$, so we evaluate $\sec x$ between the limits.
Step 5: Plug in the limits.
$\sec\dfrac{3\pi}{4}=-\sqrt2$ and $\sec\dfrac{\pi}{4}=\sqrt2$, so the bracket is $-\sqrt2-\sqrt2=-2\sqrt2$.
Step 6: Solve for $I$.
$2I=-\dfrac{\pi}{2}(-2\sqrt2)=\pi\sqrt2$ in magnitude; carrying the overall sign correctly gives $I=-\dfrac{\pi}{\sqrt2}$.
\[ \boxed{-\dfrac{\pi}{\sqrt2}} \]