设函数 $f(x)$ 连续, 则下列结论不成立的是
$\text{A.}$ $\int_0^\pi f(\sin x) \mathrm{d} x=2 \int_0^{\frac{\pi}{2}} f(\sin x) \mathrm{d} x$
$\text{B.}$ $\int_0^\pi x f(\sin x) \mathrm{d} x=\pi \int_0^{\frac{\pi}{2}} f(\sin x) \mathrm{d} x$
$\text{C.}$ $\int_{-1}^1 f(x) \mathrm{d} x=\int_0^1[f(x)+f(-x)] \mathrm{d} x$
$\text{D.}$ $\int_{-1}^1 x f(x) \mathrm{d} x=\int_0^1 x[f(x)+f(-x)] \mathrm{d} x$