Integralen av $$$4 \tan^{3}{\left(x \right)}$$$

Bestäm $$$\int 4 \tan^{3}{\left(x \right)}\, dx$$$.

Tillämpa konstantfaktorregeln $$$\int c f{\left(x \right)}\, dx = c \int f{\left(x \right)}\, dx$$$ med $$$c=4$$$ och $$$f{\left(x \right)} = \tan^{3}{\left(x \right)}$$$:

$${\color{red}{\int{4 \tan^{3}{\left(x \right)} d x}}} = {\color{red}{\left(4 \int{\tan^{3}{\left(x \right)} d x}\right)}}$$

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

Då gäller $$$x=\operatorname{atan}{\left(u \right)}$$$ och $$$dx=\left(\operatorname{atan}{\left(u \right)}\right)^{\prime }du = \frac{du}{u^{2} + 1}$$$ (stegen kan ses »).

Integralen blir

$$4 {\color{red}{\int{\tan^{3}{\left(x \right)} d x}}} = 4 {\color{red}{\int{\frac{u^{3}}{u^{2} + 1} d u}}}$$

Eftersom graden hos täljaren inte är mindre än graden hos nämnaren, utför polynomdivision (stegen kan ses »):

$$4 {\color{red}{\int{\frac{u^{3}}{u^{2} + 1} d u}}} = 4 {\color{red}{\int{\left(u - \frac{u}{u^{2} + 1}\right)d u}}}$$

Integrera termvis:

$$4 {\color{red}{\int{\left(u - \frac{u}{u^{2} + 1}\right)d u}}} = 4 {\color{red}{\left(\int{u d u} - \int{\frac{u}{u^{2} + 1} d u}\right)}}$$

Tillämpa potensregeln $$$\int u^{n}\, du = \frac{u^{n + 1}}{n + 1}$$$ $$$\left(n \neq -1 \right)$$$ med $$$n=1$$$:

$$- 4 \int{\frac{u}{u^{2} + 1} d u} + 4 {\color{red}{\int{u d u}}}=- 4 \int{\frac{u}{u^{2} + 1} d u} + 4 {\color{red}{\frac{u^{1 + 1}}{1 + 1}}}=- 4 \int{\frac{u}{u^{2} + 1} d u} + 4 {\color{red}{\left(\frac{u^{2}}{2}\right)}}$$

Låt $$$v=u^{2} + 1$$$ vara.

Då $$$dv=\left(u^{2} + 1\right)^{\prime }du = 2 u du$$$ (stegen kan ses »), och vi har att $$$u du = \frac{dv}{2}$$$.

Integralen blir

$$2 u^{2} - 4 {\color{red}{\int{\frac{u}{u^{2} + 1} d u}}} = 2 u^{2} - 4 {\color{red}{\int{\frac{1}{2 v} d v}}}$$

Tillämpa konstantfaktorregeln $$$\int c f{\left(v \right)}\, dv = c \int f{\left(v \right)}\, dv$$$ med $$$c=\frac{1}{2}$$$ och $$$f{\left(v \right)} = \frac{1}{v}$$$:

$$2 u^{2} - 4 {\color{red}{\int{\frac{1}{2 v} d v}}} = 2 u^{2} - 4 {\color{red}{\left(\frac{\int{\frac{1}{v} d v}}{2}\right)}}$$

Integralen av $$$\frac{1}{v}$$$ är $$$\int{\frac{1}{v} d v} = \ln{\left(\left|{v}\right| \right)}$$$:

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

Kom ihåg att $$$v=u^{2} + 1$$$:

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

Kom ihåg att $$$u=\tan{\left(x \right)}$$$:

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

Alltså,

$$\int{4 \tan^{3}{\left(x \right)} d x} = - 2 \ln{\left(\tan^{2}{\left(x \right)} + 1 \right)} + 2 \tan^{2}{\left(x \right)}$$

Lägg till integrationskonstanten:

$$\int{4 \tan^{3}{\left(x \right)} d x} = - 2 \ln{\left(\tan^{2}{\left(x \right)} + 1 \right)} + 2 \tan^{2}{\left(x \right)}+C$$

$$$\int 4 \tan^{3}{\left(x \right)}\, dx = \left(- 2 \ln\left(\tan^{2}{\left(x \right)} + 1\right) + 2 \tan^{2}{\left(x \right)}\right) + C$$$A