曲线 $y=\int_0^x \tan t \mathrm{~d} t\left(0 \leqslant x \leqslant \frac{\pi}{3}\right)$ 的弧长 $s=$
【答案】 $\ln (2+\sqrt{3})$

【解析】
$$
\begin{aligned}
s &=\int_L \mathrm{~d} s=\int_0^{\frac{\pi}{3}} \sqrt{1+\left(\frac{\mathrm{d} y}{\mathrm{~d} x}\right)^2} \mathrm{~d} x=\int_0^{\frac{\pi}{3}} \sqrt{1+(\tan x)^2} \mathrm{~d} x=\int_0^{\frac{\pi}{3}} \sec x \mathrm{~d} x \\
&=\left.\ln |\sec x+\tan x|\right|_0 ^{\frac{\pi}{3}}=\ln (2+\sqrt{3})
\end{aligned}
$$
系统推荐