Đáp án:

$B.\ \dfrac{8 - 5\sqrt2}{3}$

Giải thích các bước giải:

$\quad \displaystyle\int\limits_0^{\pi}(\sin x - x).f'\left(\dfrac x2\right)dx= -2\pi$

Đặt $t =\dfrac x2$

$\Rightarrow dt = \dfrac12dx$

Đổi cận:

$\begin{array}{c|cc}x&0&&\pi\\\hline t&0&&\dfrac{\pi}{2}\end{array}$

Ta được:

$\quad \displaystyle\int\limits_0^{\tfrac{\pi}{2}}(\sin2t - 2t)f'(t)2dt = - 2\pi\qquad (*)$

Đặt $\begin{cases}u =\sin2t - 2t\\dv = f'(t)dt\end{cases}\Rightarrow \begin{cases}du = 2\cos2t - 2\\v = f(t)\end{cases}$

Ta được:

$(*)\Leftrightarrow 2f(t)(\sin2t - 2t)\Bigg|_0^{\tfrac{\pi}{2}} - 2\displaystyle\int\limits_0^{\tfrac{\pi}{2}}f(t)(2\cos2t - 2)dt= - 2\pi$

$\Leftrightarrow - 8\pi + 8\displaystyle\int\limits_0^{\tfrac{\pi}{2}}f(t)\sin^2tdt = - 2\pi$

$\Leftrightarrow -8\displaystyle\int\limits_0^{\tfrac{\pi}{2}}f(t)\sin^2tdt =- 6\pi$

$\Leftrightarrow -8\displaystyle\int\limits_0^{\tfrac{\pi}{2}}f(x)\sin^2xdx=- 6\pi\qquad (1)$

Ta lại có:

$\displaystyle\int\limits_0^{\tfrac{\pi}{2}}f^2(x)dx = 3\pi\qquad\quad (2)$

$\displaystyle\int\limits_0^{\tfrac{\pi}{2}}16\sin^4xdx = 3\pi\qquad (3)$

Cộng vế theo vế $(1),(2),(3)$ ta được:

$\quad \displaystyle\int\limits_0^{\tfrac{\pi}{2}}[f^2(x) - 8f(x)\sin^2x + 16\sin^4x]dx = 0$

$\Leftrightarrow \displaystyle\int\limits_0^{\tfrac{\pi}{2}}[f(x) - 4\sin^2x]^2dx = 0$

$\Leftrightarrow f(x) = 4\sin^2x$

Khi đó:

$\quad \displaystyle\int\limits_0^{\tfrac{\pi}{4}}f(x)\sin xdx$

$= \displaystyle\int\limits_0^{\tfrac{\pi}{4}}4\sin^3xdx$

$= \dfrac{8 - 5\sqrt2}{3}$