Funktion $$$\tan{\left(\theta \right)}$$$ integraali

Määritä $$$\int \tan{\left(\theta \right)}\, d\theta$$$.

Kirjoita tangentti uudelleen muotoon $$$\tan\left(\theta\right)=\frac{\sin\left(\theta\right)}{\cos\left(\theta\right)}$$$:

$${\color{red}{\int{\tan{\left(\theta \right)} d \theta}}} = {\color{red}{\int{\frac{\sin{\left(\theta \right)}}{\cos{\left(\theta \right)}} d \theta}}}$$

Olkoon $$$u=\cos{\left(\theta \right)}$$$.

Tällöin $$$du=\left(\cos{\left(\theta \right)}\right)^{\prime }d\theta = - \sin{\left(\theta \right)} d\theta$$$ (vaiheet ovat nähtävissä ») ja saamme, että $$$\sin{\left(\theta \right)} d\theta = - du$$$.

Integraali muuttuu muotoon

$${\color{red}{\int{\frac{\sin{\left(\theta \right)}}{\cos{\left(\theta \right)}} d \theta}}} = {\color{red}{\int{\left(- \frac{1}{u}\right)d u}}}$$

Sovella vakiokertoimen sääntöä $$$\int c f{\left(u \right)}\, du = c \int f{\left(u \right)}\, du$$$ käyttäen $$$c=-1$$$ ja $$$f{\left(u \right)} = \frac{1}{u}$$$:

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

Funktion $$$\frac{1}{u}$$$ integraali on $$$\int{\frac{1}{u} d u} = \ln{\left(\left|{u}\right| \right)}$$$:

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

Muista, että $$$u=\cos{\left(\theta \right)}$$$:

$$- \ln{\left(\left|{{\color{red}{u}}}\right| \right)} = - \ln{\left(\left|{{\color{red}{\cos{\left(\theta \right)}}}}\right| \right)}$$

Näin ollen,

$$\int{\tan{\left(\theta \right)} d \theta} = - \ln{\left(\left|{\cos{\left(\theta \right)}}\right| \right)}$$

Lisää integrointivakio:

$$\int{\tan{\left(\theta \right)} d \theta} = - \ln{\left(\left|{\cos{\left(\theta \right)}}\right| \right)}+C$$

$$$\int \tan{\left(\theta \right)}\, d\theta = - \ln\left(\left|{\cos{\left(\theta \right)}}\right|\right) + C$$$A