Integralen av $$$\frac{1}{\sin^{4}{\left(x \right)} \cos^{4}{\left(x \right)}}$$$

Bestäm $$$\int \frac{1}{\sin^{4}{\left(x \right)} \cos^{4}{\left(x \right)}}\, dx$$$.

Multiplicera täljaren och nämnaren med $$$\frac{1}{\cos^{4}{\left(x \right)}}$$$ och omvandla $$$\frac{\cos^{4}{\left(x \right)}}{\sin^{4}{\left(x \right)}}$$$ till $$$\frac{1}{\tan^{4}{\left(x \right)}}$$$:

$${\color{red}{\int{\frac{1}{\sin^{4}{\left(x \right)} \cos^{4}{\left(x \right)}} d x}}} = {\color{red}{\int{\frac{1}{\cos^{8}{\left(x \right)} \tan^{4}{\left(x \right)}} d x}}}$$

Bryt ut två cosinus och skriv om dem i termer av sekant med hjälp av formeln $$$\frac{1}{\cos^{2}{\left(x \right)}}=\sec^{2}{\left(x \right)}$$$:

$${\color{red}{\int{\frac{1}{\cos^{8}{\left(x \right)} \tan^{4}{\left(x \right)}} d x}}} = {\color{red}{\int{\frac{\sec^{2}{\left(x \right)}}{\cos^{6}{\left(x \right)} \tan^{4}{\left(x \right)}} d x}}}$$

Skriv om cosinus i termer av tangens med hjälp av formeln $$$\cos^{2}{\left(x \right)}=\frac{1}{\tan^{2}{\left(x \right)} + 1}$$$:

$${\color{red}{\int{\frac{\sec^{2}{\left(x \right)}}{\cos^{6}{\left(x \right)} \tan^{4}{\left(x \right)}} d x}}} = {\color{red}{\int{\frac{\left(\tan^{2}{\left(x \right)} + 1\right)^{3} \sec^{2}{\left(x \right)}}{\tan^{4}{\left(x \right)}} d x}}}$$

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

Då $$$du=\left(\tan{\left(x \right)}\right)^{\prime }dx = \sec^{2}{\left(x \right)} dx$$$ (stegen kan ses »), och vi har att $$$\sec^{2}{\left(x \right)} dx = du$$$.

Alltså,

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

Expand the expression:

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

Integrera termvis:

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

Tillämpa konstantregeln $$$\int c\, du = c u$$$ med $$$c=3$$$:

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

Tillämpa potensregeln $$$\int u^{n}\, du = \frac{u^{n + 1}}{n + 1}$$$ $$$\left(n \neq -1 \right)$$$ med $$$n=-4$$$:

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

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

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

Tillämpa konstantfaktorregeln $$$\int c f{\left(u \right)}\, du = c \int f{\left(u \right)}\, du$$$ med $$$c=3$$$ och $$$f{\left(u \right)} = \frac{1}{u^{2}}$$$:

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

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

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

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

$$- \frac{{\color{red}{u}}^{-3}}{3} - 3 {\color{red}{u}}^{-1} + 3 {\color{red}{u}} + \frac{{\color{red}{u}}^{3}}{3} = - \frac{{\color{red}{\tan{\left(x \right)}}}^{-3}}{3} - 3 {\color{red}{\tan{\left(x \right)}}}^{-1} + 3 {\color{red}{\tan{\left(x \right)}}} + \frac{{\color{red}{\tan{\left(x \right)}}}^{3}}{3}$$

Alltså,

$$\int{\frac{1}{\sin^{4}{\left(x \right)} \cos^{4}{\left(x \right)}} d x} = \frac{\tan^{3}{\left(x \right)}}{3} + 3 \tan{\left(x \right)} - \frac{3}{\tan{\left(x \right)}} - \frac{1}{3 \tan^{3}{\left(x \right)}}$$

Lägg till integrationskonstanten:

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

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