Integralen av $$$8 \tan^{3}{\left(x \right)} \sec{\left(x \right)}$$$

Bestäm $$$\int 8 \tan^{3}{\left(x \right)} \sec{\left(x \right)}\, dx$$$.

Tillämpa konstantfaktorregeln $$$\int c f{\left(x \right)}\, dx = c \int f{\left(x \right)}\, dx$$$ med $$$c=8$$$ och $$$f{\left(x \right)} = \tan^{3}{\left(x \right)} \sec{\left(x \right)}$$$:

$${\color{red}{\int{8 \tan^{3}{\left(x \right)} \sec{\left(x \right)} d x}}} = {\color{red}{\left(8 \int{\tan^{3}{\left(x \right)} \sec{\left(x \right)} d x}\right)}}$$

Bryt ut en tangens och skriv allt annat i termer av sekanten, med hjälp av formeln $$$\tan^2\left(x \right)=\sec^2\left(x \right)-1$$$:

$$8 {\color{red}{\int{\tan^{3}{\left(x \right)} \sec{\left(x \right)} d x}}} = 8 {\color{red}{\int{\left(\sec^{2}{\left(x \right)} - 1\right) \tan{\left(x \right)} \sec{\left(x \right)} d x}}}$$

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

Då $$$du=\left(\sec{\left(x \right)}\right)^{\prime }dx = \tan{\left(x \right)} \sec{\left(x \right)} dx$$$ (stegen kan ses »), och vi har att $$$\tan{\left(x \right)} \sec{\left(x \right)} dx = du$$$.

Alltså,

$$8 {\color{red}{\int{\left(\sec^{2}{\left(x \right)} - 1\right) \tan{\left(x \right)} \sec{\left(x \right)} d x}}} = 8 {\color{red}{\int{\left(u^{2} - 1\right)d u}}}$$

Integrera termvis:

$$8 {\color{red}{\int{\left(u^{2} - 1\right)d u}}} = 8 {\color{red}{\left(- \int{1 d u} + \int{u^{2} d u}\right)}}$$

Tillämpa konstantregeln $$$\int c\, du = c u$$$ med $$$c=1$$$:

$$8 \int{u^{2} d u} - 8 {\color{red}{\int{1 d u}}} = 8 \int{u^{2} d u} - 8 {\color{red}{u}}$$

Tillämpa potensregeln $$$\int u^{n}\, du = \frac{u^{n + 1}}{n + 1}$$$ $$$\left(n \neq -1 \right)$$$ med $$$n=2$$$:

$$- 8 u + 8 {\color{red}{\int{u^{2} d u}}}=- 8 u + 8 {\color{red}{\frac{u^{1 + 2}}{1 + 2}}}=- 8 u + 8 {\color{red}{\left(\frac{u^{3}}{3}\right)}}$$

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

$$- 8 {\color{red}{u}} + \frac{8 {\color{red}{u}}^{3}}{3} = - 8 {\color{red}{\sec{\left(x \right)}}} + \frac{8 {\color{red}{\sec{\left(x \right)}}}^{3}}{3}$$

Alltså,

$$\int{8 \tan^{3}{\left(x \right)} \sec{\left(x \right)} d x} = \frac{8 \sec^{3}{\left(x \right)}}{3} - 8 \sec{\left(x \right)}$$

Förenkla:

$$\int{8 \tan^{3}{\left(x \right)} \sec{\left(x \right)} d x} = \frac{8 \left(\sec^{2}{\left(x \right)} - 3\right) \sec{\left(x \right)}}{3}$$

Lägg till integrationskonstanten:

$$\int{8 \tan^{3}{\left(x \right)} \sec{\left(x \right)} d x} = \frac{8 \left(\sec^{2}{\left(x \right)} - 3\right) \sec{\left(x \right)}}{3}+C$$

$$$\int 8 \tan^{3}{\left(x \right)} \sec{\left(x \right)}\, dx = \frac{8 \left(\sec^{2}{\left(x \right)} - 3\right) \sec{\left(x \right)}}{3} + C$$$A