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.} $ $\int_{-1}^1 x f(x) \mathrm{d} x=\int_0^1 x[f(x)+f(-x)] \mathrm{d} x$
【答案】 D

【解析】 【解】
$$
\begin{aligned}
\int_0^\pi f(\sin x) \mathrm{d} x= & \int_0^{\frac{\pi}{2}} f(\sin x) \mathrm{d} x+\int_{\frac{\pi}{2}}^\pi f(\sin x) \mathrm{d} x \stackrel{\text { 第 } 2 \text { 个积分中 }}{\frac{\text { 令 } x=\pi-t}{2}} \int_0^{\frac{\pi}{2}} f(\sin x) \mathrm{d} x \\
& +\int_0^{\frac{\pi}{2}} f(\sin t) \mathrm{d} t=2 \int_0^{\frac{\pi}{2}} f(\sin x) \mathrm{d} x .
\end{aligned}
$$
$\mathrm{A}$ 成立.

$$
\begin{aligned}
\int_0^\pi x f(\sin x) \mathrm{d} x & \stackrel{x=\pi-t}{=} \int_0^\pi(\pi-t) f(\sin t) \mathrm{d} t=\pi \int_0^\pi f(\sin t) \mathrm{d} t-\int_0^\pi t f(\sin t) \mathrm{d} t \\
& =\pi \int_0^\pi f(\sin x) \mathrm{d} x-\int_0^\pi x f(\sin x) \mathrm{d} x \\
& =\frac{\pi}{2} \int_0^\pi f(\sin x) \mathrm{d} x=\pi \int_0^{\frac{\pi}{2}} f(\sin x) \mathrm{d} x .
\end{aligned}
$$
$\mathrm{B}$ 成立.
$$
\int_{-1}^1 f(x) \mathrm{d} x \stackrel{x=-t}{=} \int_{-1}^1 f(-t) \mathrm{d} t=\int_{-1}^1 f(-x) \mathrm{d} x=\frac{1}{2} \int_{-1}^1[f(x)+f(-x)] \mathrm{d} x
$$
奇偶性 $\int_0^1[f(x)+f(-x)] \mathrm{d} x$.
$\mathrm{C}$ 成立.
$$
\begin{aligned}
& \int_{-1}^1 x f(x) \mathrm{d} x \stackrel{x=-t}{=}-\int_{-1}^1 t f(-t) \mathrm{d} t=-\int_{-1}^1 x f(-x) \mathrm{d} x=\frac{1}{2} \int_{-1}^1 x[f(x)-f(-x)] \mathrm{d} x \\
& \stackrel{\text { 奇偶性 }}{=} \int_0^1 x[f(x)-f(-x)] \mathrm{d} x \text {. } \\
&
\end{aligned}
$$
$\mathrm{D}$ 不成立.
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