$$$\frac{1}{\cos^{2}{\left(t \right)} \tan{\left(t \right)}}$$$ 的積分

求$$$\int \frac{1}{\cos^{2}{\left(t \right)} \tan{\left(t \right)}}\, dt$$$。

重寫被積函數:

$${\color{red}{\int{\frac{1}{\cos^{2}{\left(t \right)} \tan{\left(t \right)}} d t}}} = {\color{red}{\int{\frac{\sec^{2}{\left(t \right)}}{\tan{\left(t \right)}} d t}}}$$

令 $$$u=\tan{\left(t \right)}$$$。

則 $$$du=\left(\tan{\left(t \right)}\right)^{\prime }dt = \sec^{2}{\left(t \right)} dt$$$ (步驟見»)，並可得 $$$\sec^{2}{\left(t \right)} dt = du$$$。

該積分變為

$${\color{red}{\int{\frac{\sec^{2}{\left(t \right)}}{\tan{\left(t \right)}} d t}}} = {\color{red}{\int{\frac{1}{u} d u}}}$$

$$$\frac{1}{u}$$$ 的積分是 $$$\int{\frac{1}{u} d u} = \ln{\left(\left|{u}\right| \right)}$$$：

$${\color{red}{\int{\frac{1}{u} d u}}} = {\color{red}{\ln{\left(\left|{u}\right| \right)}}}$$

回顧一下 $$$u=\tan{\left(t \right)}$$$：

$$\ln{\left(\left|{{\color{red}{u}}}\right| \right)} = \ln{\left(\left|{{\color{red}{\tan{\left(t \right)}}}}\right| \right)}$$

因此，

$$\int{\frac{1}{\cos^{2}{\left(t \right)} \tan{\left(t \right)}} d t} = \ln{\left(\left|{\tan{\left(t \right)}}\right| \right)}$$

加上積分常數：

$$\int{\frac{1}{\cos^{2}{\left(t \right)} \tan{\left(t \right)}} d t} = \ln{\left(\left|{\tan{\left(t \right)}}\right| \right)}+C$$

$$$\int \frac{1}{\cos^{2}{\left(t \right)} \tan{\left(t \right)}}\, dt = \ln\left(\left|{\tan{\left(t \right)}}\right|\right) + C$$$A