$\lim _{t \rightarrow 0^{+}} \frac{1}{t^3} \int_0^{\frac{\pi}{4}} \mathrm{~d} \theta \int_0^{\frac{t}{\cos \theta}} \frac{\sin \left(r^2 \sin \theta \cos \theta\right)}{\sin \theta} \mathrm{d} r=$
【答案】 $\frac{1}{3}$

【解析】 【解】
$$
\begin{aligned}
\lim _{t \rightarrow 0^{+}} \frac{1}{t^3} \int_0^{\frac{\pi}{4}} \mathrm{~d} \theta \int_0^{\frac{1}{\cos \theta}} \frac{\sin \left(r^2 \sin \theta \cos \theta\right)}{\sin \theta} \mathrm{d} r & =\lim _{t \rightarrow 0^{+}} \frac{\int_0^t \mathrm{~d} x \int_0^x \frac{\sin (x y)}{y} \mathrm{~d} y}{t^3} \\
& =\lim _{t \rightarrow 0^{+}} \frac{\int_0^t \frac{\sin (t y)}{y} \mathrm{~d} y}{3 t^2} \frac{u=t y}{=} \lim _{t \rightarrow 0^{+}} \frac{\int_0^{t^2} \frac{\sin u}{u} \mathrm{~d} u}{3 t^2} \\
& =\lim _{t \rightarrow 0^{+}} \frac{\frac{\sin \left(t^2\right)}{t^2} \cdot 2 t}{6 t}=\frac{1}{3}
\end{aligned}
$$
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