(a) 求 $-\ln\cos x$ 的麦克劳林级数中 $x^2$ 项的系数。

(b) 求曲线 $y=-\ln\cos x$ 从 $x=0$ 的点到 $x=\frac{1}{4}\pi$ 的点之间的弧长。

Answer all questions. You should use a calculator where appropriate.

You must show all necessary working clearly; no marks will be given for unsupported answers from a calculator.

Give non-exact numerical answers correct to 3 significant figures, or 1 decimal place for angles in degrees, unless a different level of accuracy is specified in the question.

请回答所有题目。适当时应使用计算器。必须清楚展示所有必要的计算过程；仅给出计算器答案而没有过程，不予评分。除非题目另有精度要求，非精确数值答案保留 3 位有效数字，以度为单位的角度保留 1 位小数。

二次项的系数为 $f''(0)/2!$。两次求导后，只需在 $x=0$ 处算正割的平方。

For $f(x)=-\ln(\cos x)$, we have $f'(x)=\tan x$ and $f''(x)=\sec^2x$. Using $f(x)=f(0)+f'(0)x+\frac12f''(0)x^2+\cdots$, the coefficient is $f''(0)/2!$. Since $f''(0)=\sec^20=1$, it is $\boxed{\frac12}$.

也可先展开 $\cos x$，再把它代入 $-\ln(1+u)$。这里 $u$ 的最低次项已经是二次项。

Since $\cos x=1-\frac12x^2+O(x^4)$, we have $u=\cos x-1=-\frac12x^2+O(x^4)$. The standard expansion is $\ln(1+u)=u-u^2/2+\cdots$. Therefore $-\ln(1+u)=-u+O(u^2)=\frac12x^2+O(x^4)$, giving the coefficient $\frac12$.

弧长因子是 $\sqrt{1+\tan^2x}$。在给定区间内 $\sec x>0$，可直接积分正割。

On $0\le x\le\pi/4$, we have $\sec x>0$, so $\sqrt{1+\tan^2x}=\sqrt{\sec^2x}=\sec x$. Using the arc-length formula $s=\int\sqrt{1+(y\prime)^2}\,\mathrm{d}x$, the length is