Funktion $$$\frac{1}{\tan{\left(x \right)}}$$$ integraali

Määritä $$$\int \frac{1}{\tan{\left(x \right)}}\, dx$$$.

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

Tällöin $$$x=\operatorname{atan}{\left(u \right)}$$$ ja $$$dx=\left(\operatorname{atan}{\left(u \right)}\right)^{\prime }du = \frac{du}{u^{2} + 1}$$$ (vaiheet ovat nähtävissä »).

Integraali voidaan kirjoittaa muotoon

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

Olkoon $$$v=u^{2} + 1$$$.

Tällöin $$$dv=\left(u^{2} + 1\right)^{\prime }du = 2 u du$$$ (vaiheet ovat nähtävissä ») ja saamme, että $$$u du = \frac{dv}{2}$$$.

Näin ollen,

$${\color{red}{\int{\frac{1}{u \left(u^{2} + 1\right)} d u}}} = {\color{red}{\int{\frac{1}{2 v \left(v - 1\right)} d v}}}$$

Sovella vakiokertoimen sääntöä $$$\int c f{\left(v \right)}\, dv = c \int f{\left(v \right)}\, dv$$$ käyttäen $$$c=\frac{1}{2}$$$ ja $$$f{\left(v \right)} = \frac{1}{v \left(v - 1\right)}$$$:

$${\color{red}{\int{\frac{1}{2 v \left(v - 1\right)} d v}}} = {\color{red}{\left(\frac{\int{\frac{1}{v \left(v - 1\right)} d v}}{2}\right)}}$$

Suorita osamurtokehittely (vaiheet voidaan nähdä kohdassa »):

$$\frac{{\color{red}{\int{\frac{1}{v \left(v - 1\right)} d v}}}}{2} = \frac{{\color{red}{\int{\left(\frac{1}{v - 1} - \frac{1}{v}\right)d v}}}}{2}$$

Integroi termi kerrallaan:

$$\frac{{\color{red}{\int{\left(\frac{1}{v - 1} - \frac{1}{v}\right)d v}}}}{2} = \frac{{\color{red}{\left(- \int{\frac{1}{v} d v} + \int{\frac{1}{v - 1} d v}\right)}}}{2}$$

Olkoon $$$w=v - 1$$$.

Tällöin $$$dw=\left(v - 1\right)^{\prime }dv = 1 dv$$$ (vaiheet ovat nähtävissä ») ja saamme, että $$$dv = dw$$$.

Siis,

$$- \frac{\int{\frac{1}{v} d v}}{2} + \frac{{\color{red}{\int{\frac{1}{v - 1} d v}}}}{2} = - \frac{\int{\frac{1}{v} d v}}{2} + \frac{{\color{red}{\int{\frac{1}{w} d w}}}}{2}$$

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

$$- \frac{\int{\frac{1}{v} d v}}{2} + \frac{{\color{red}{\int{\frac{1}{w} d w}}}}{2} = - \frac{\int{\frac{1}{v} d v}}{2} + \frac{{\color{red}{\ln{\left(\left|{w}\right| \right)}}}}{2}$$

Muista, että $$$w=v - 1$$$:

$$\frac{\ln{\left(\left|{{\color{red}{w}}}\right| \right)}}{2} - \frac{\int{\frac{1}{v} d v}}{2} = \frac{\ln{\left(\left|{{\color{red}{\left(v - 1\right)}}}\right| \right)}}{2} - \frac{\int{\frac{1}{v} d v}}{2}$$

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

$$\frac{\ln{\left(\left|{v - 1}\right| \right)}}{2} - \frac{{\color{red}{\int{\frac{1}{v} d v}}}}{2} = \frac{\ln{\left(\left|{v - 1}\right| \right)}}{2} - \frac{{\color{red}{\ln{\left(\left|{v}\right| \right)}}}}{2}$$

Muista, että $$$v=u^{2} + 1$$$:

$$\frac{\ln{\left(\left|{-1 + {\color{red}{v}}}\right| \right)}}{2} - \frac{\ln{\left(\left|{{\color{red}{v}}}\right| \right)}}{2} = \frac{\ln{\left(\left|{-1 + {\color{red}{\left(u^{2} + 1\right)}}}\right| \right)}}{2} - \frac{\ln{\left(\left|{{\color{red}{\left(u^{2} + 1\right)}}}\right| \right)}}{2}$$

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

$$- \frac{\ln{\left(1 + {\color{red}{u}}^{2} \right)}}{2} + \frac{\ln{\left({\color{red}{u}}^{2} \right)}}{2} = - \frac{\ln{\left(1 + {\color{red}{\tan{\left(x \right)}}}^{2} \right)}}{2} + \frac{\ln{\left({\color{red}{\tan{\left(x \right)}}}^{2} \right)}}{2}$$

Näin ollen,

$$\int{\frac{1}{\tan{\left(x \right)}} d x} = - \frac{\ln{\left(\tan^{2}{\left(x \right)} + 1 \right)}}{2} + \frac{\ln{\left(\tan^{2}{\left(x \right)} \right)}}{2}$$

Sievennä:

$$\int{\frac{1}{\tan{\left(x \right)}} d x} = - \frac{\ln{\left(\tan^{2}{\left(x \right)} + 1 \right)}}{2} + \ln{\left(\tan{\left(x \right)} \right)}$$

Lisää integrointivakio:

$$\int{\frac{1}{\tan{\left(x \right)}} d x} = - \frac{\ln{\left(\tan^{2}{\left(x \right)} + 1 \right)}}{2} + \ln{\left(\tan{\left(x \right)} \right)}+C$$

$$$\int \frac{1}{\tan{\left(x \right)}}\, dx = \left(- \frac{\ln\left(\tan^{2}{\left(x \right)} + 1\right)}{2} + \ln\left(\tan{\left(x \right)}\right)\right) + C$$$A