Funktion $$$\frac{x}{\cos^{2}{\left(x \right)}}$$$ integraali

Määritä $$$\int \frac{x}{\cos^{2}{\left(x \right)}}\, dx$$$.

Integraalin $$$\int{\frac{x}{\cos^{2}{\left(x \right)}} d x}$$$ kohdalla käytä osittaisintegrointia $$$\int \operatorname{u} \operatorname{dv} = \operatorname{u}\operatorname{v} - \int \operatorname{v} \operatorname{du}$$$.

Olkoon $$$\operatorname{u}=x$$$ ja $$$\operatorname{dv}=\frac{dx}{\cos^{2}{\left(x \right)}}$$$.

Tällöin $$$\operatorname{du}=\left(x\right)^{\prime }dx=1 dx$$$ (vaiheet ovat nähtävissä ») ja $$$\operatorname{v}=\int{\frac{1}{\cos^{2}{\left(x \right)}} d x}=\tan{\left(x \right)}$$$ (vaiheet ovat nähtävissä »).

Siis,

$${\color{red}{\int{\frac{x}{\cos^{2}{\left(x \right)}} d x}}}={\color{red}{\left(x \cdot \tan{\left(x \right)}-\int{\tan{\left(x \right)} \cdot 1 d x}\right)}}={\color{red}{\left(x \tan{\left(x \right)} - \int{\tan{\left(x \right)} d x}\right)}}$$

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

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

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

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

Integraali muuttuu muotoon

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

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

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

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

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

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

$$x \tan{\left(x \right)} + \ln{\left(\left|{{\color{red}{u}}}\right| \right)} = x \tan{\left(x \right)} + \ln{\left(\left|{{\color{red}{\cos{\left(x \right)}}}}\right| \right)}$$

Näin ollen,

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

Lisää integrointivakio:

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

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