# 2011年考研数二第06题解析

## 题目

$$(A) I < J < K$$

$$(B) I < K < J$$

$$(C) J < I < K$$

$$(D) K < J < I$$

## 解析

$$\sin x \rightarrow 0;$$

$$\cot x \rightarrow \infty;$$

$$\cos x \rightarrow 1.$$

$$\sin x \rightarrow \frac{\sqrt{2}}{2};$$

$$\cot x \rightarrow 1;$$

$$\cos x \rightarrow \frac{\sqrt{2}}{2}.$$

$$\ln \sin x \rightarrow – \infty;$$

$$\ln \cot x \rightarrow + \infty.$$

$$\int_{0}^{\frac{\pi}{4}} \ln \sin x dx =$$

$$x \cdot \ln \sin x |_{0}^{\frac{\pi}{4}} – \int_{0}^{\frac{\pi}{4}}x d(\ln \sin x) =$$
$$x \cdot \ln \sin x |_{0}^{\frac{\pi}{4}} – \int_{0}^{\frac{\pi}{4}} x \cdot \frac{\cos x}{\sin x} dx.$$

(1)

$$x \cdot \ln \sin x =$$

$$\frac{x}{\sin x} \cdot \sin x \cdot \ln \sin x =$$

$$1 \cdot \sin x \cdot \ln \sin x \Rightarrow$$

$$令 t = \sin x, t \rightarrow 0, 则：$$

$$\sin x \cdot \ln \sin x =$$

$$t \cdot \ln t =$$

$$\frac{\ln t}{\frac{1}{t}} =$$

$$\frac{(\ln t)^{‘}}{(\frac{1}{t})^{‘}} =$$

$$\frac{\frac{1}{t}}{\frac{1}{-t^{2}}} = -t = 0.$$

(2)

$$x \cdot \frac{\cos x}{\sin x} =$$

$$\frac{x}{\sin x} \cdot \cos x =$$

$$1 \cdot \cos x = \cos x = 1.$$

$\int_{0}^{\frac{\pi}{4}} x \cdot \ln \sin x$ 和 $\int_{0}^{\frac{\pi}{4}} x \cdot \frac{\cos x}{\sin x}$ 均为收敛积分，故：

$\int_{0}^{\frac{\pi}{4}} \ln \sin x dx$ 是一个收敛积分。

$$\int_{0}^{\frac{\pi}{4}} \ln \cot x dx =$$

$$\int_{0}^{\frac{\pi}{4}} \ln (\frac{\cos x}{\sin x}) dx =$$

$$\int_{0}^{\frac{\pi}{4}} (\ln \cos x – \ln \sin x) dx =$$

$$\int_{0}^{\frac{\pi}{4}} \ln \cos x dx – \int_{0}^{\frac{\pi}{4}} \ln \sin x dx \Rightarrow$$

$\int_{0}^{\frac{\pi}{4}} \ln \cot x dx$ 是一个收敛积分。

$$\sin x < \cos x < \cot x.$$

$$I < K < J.$$