(arcsin^2x+x^2tanx)/sqrt(1-x^2) 在x属于[-1,1]上的定积分

我们可以将被积函数拆开，分别考虑：

$$
\begin{aligned}
\int_{-1}^{1} \frac{\arcsin^2 x}{\sqrt{1-x^2}} \mathrm{d}x &= 2\int_{0}^{1} \frac{\arcsin^2 x}{\sqrt{1-x^2}} \mathrm{d}x \\
\int_{-1}^{1} \frac{x^2 \tan x}{\sqrt{1-x^2}} \mathrm{d}x &= \int_{-1}^{1} \frac{x^2}{\sqrt{1-x^2}} \cdot \frac{\sin x}{\cos x} \mathrm{d}x \\
&= -\int_{-1}^{1} x^2 \mathrm{d}(\sqrt{1-x^2}) \\
&= 2\int_{0}^{1} (1-t^2) \mathrm{d}t, \quad \text{令} t = \sqrt{1-x^2}
\end{aligned}
$$

因此，原式可以表示为：

$$
\begin{aligned}
\int_{-1}^{1} \frac{\arcsin^2 x}{\sqrt{1-x^2}} \mathrm{d}x + \int_{-1}^{1} \frac{x^2 \tan x}{\sqrt{1-x^2}} \mathrm{d}x &= 2\int_{0}^{1} \frac{\arcsin^2 x}{\sqrt{1-x^2}} \mathrm{d}x + 2\int_{0}^{1} (1-t^2) \mathrm{d}t \\
&= 2\int_{0}^{1} \left(\frac{\arcsin^2 x}{\sqrt{1-x^2}} + 1 - x^2\right) \mathrm{d}x \\
&= \left[2\left(x - \sqrt{1-x^2} + \frac{1}{3}\arcsin^3 x\right)\right]_{0}^{1} \\
&= \frac{4}{3}\pi - \frac{8}{3}
\end{aligned}
$$

因此，原式的定积分结果为 $\frac{4}{3}\pi - \frac{8}{3}$。