Integraal van $$$\frac{1}{\cos^{2}{\left(t \right)} \tan{\left(t \right)}}$$$

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

Herschrijf de integraand:

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

Zij $$$u=\tan{\left(t \right)}$$$.

Dan $$$du=\left(\tan{\left(t \right)}\right)^{\prime }dt = \sec^{2}{\left(t \right)} dt$$$ (de stappen zijn te zien »), en dan geldt dat $$$\sec^{2}{\left(t \right)} dt = du$$$.

De integraal wordt

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

De integraal van $$$\frac{1}{u}$$$ is $$$\int{\frac{1}{u} d u} = \ln{\left(\left|{u}\right| \right)}$$$:

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

We herinneren eraan dat $$$u=\tan{\left(t \right)}$$$:

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

Dus,

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

Voeg de integratieconstante toe:

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

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