# 三角函数凑微分搭配分部积分：$\int$ $\frac{1}{\cos^{3} x}$ $\mathrm{d} x$

## 一、题目

$$\int \frac{1}{\cos^{3} x} \mathrm{d} x = ?$$

## 二、解析

$$(\tan x)^{\prime} = \frac{1}{\cos^{2} x}$$

Next

$$\int \frac{1}{\cos^{3} x} \mathrm{d} x =$$

$$\int \frac{1}{\cos x} \mathrm{d} (\tan x) =$$

$$\int \sec x \mathrm{d} (\tan x) \Rightarrow$$

Next

$$\sec x \tan x – \int \tan x \mathrm{d} (\sec x) =$$

$$\sec x \tan x – \int \tan x \cdot \sec x \cdot \tan x \mathrm{d} x =$$

$$\sec x \tan x – \int \tan^{2} x \cdot \sec x \mathrm{d} x =$$

$$\sec x \tan x – \int \frac{\sin^{2} x}{\cos^{2} x} \cdot \frac{1}{\cos x} \mathrm{d} x =$$

Next

$$\sec x \tan x – \int \frac{1 – \cos^{2} x}{\cos^{3} x} \mathrm{d} x =$$

$$\sec x \tan x – \int \frac{1}{\cos^{3} x} \mathrm{d} x + \int \frac{\cos^{2} x}{\cos^{3} x} \mathrm{d} x =$$

$$\sec x \tan x – \int \frac{1}{\cos^{3} x} \mathrm{d} x + \int \frac{1}{\cos x} \mathrm{d} x =$$

$$\sec x \tan x – \int \frac{1}{\cos^{3} x} \mathrm{d} x + \int \sec x \mathrm{d} x \Rightarrow$$

Next

$$\int \frac{1}{\cos^{3} x} \mathrm{d} x = \sec x \tan x – \int \frac{1}{\cos^{3} x} \mathrm{d} x + \int \sec x \mathrm{d} x \Rightarrow$$

$$2 \int \frac{1}{\cos^{3} x} \mathrm{d} x = \sec x \tan x + \int \sec x \mathrm{d} x \Rightarrow$$

Next

$$2 \int \frac{1}{\cos^{3} x} \mathrm{d} x = \sec x \tan x + \ln |\sec x + \tan x| + C_{0} \Rightarrow$$

$$\int \frac{1}{\cos^{3} x} \mathrm{d} x = \frac{1}{2} \sec x \tan x + \frac{1}{2} \ln |\sec x + \tan x| + C.$$