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

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

Integrera termvis:

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

Skriv om kotangensen som $$$\cot\left(x\right)=\frac{\cos\left(x\right)}{\sin\left(x\right)}$$$:

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

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

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

Integralen kan omskrivas som

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

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

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

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

$$- \ln{\left(\left|{{\color{red}{u}}}\right| \right)} - \int{3 \tan^{3}{\left(3 x \right)} \cot{\left(3 x \right)} d x} + \int{\tan{\left(x \right)} d x} = - \ln{\left(\left|{{\color{red}{\sin{\left(x \right)}}}}\right| \right)} - \int{3 \tan^{3}{\left(3 x \right)} \cot{\left(3 x \right)} d x} + \int{\tan{\left(x \right)} d x}$$

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

$$- \ln{\left(\left|{\sin{\left(x \right)}}\right| \right)} + \int{\tan{\left(x \right)} d x} - {\color{red}{\int{3 \tan^{3}{\left(3 x \right)} \cot{\left(3 x \right)} d x}}} = - \ln{\left(\left|{\sin{\left(x \right)}}\right| \right)} + \int{\tan{\left(x \right)} d x} - {\color{red}{\left(3 \int{\tan^{3}{\left(3 x \right)} \cot{\left(3 x \right)} d x}\right)}}$$

Skriv om integranden:

$$- \ln{\left(\left|{\sin{\left(x \right)}}\right| \right)} + \int{\tan{\left(x \right)} d x} - 3 {\color{red}{\int{\tan^{3}{\left(3 x \right)} \cot{\left(3 x \right)} d x}}} = - \ln{\left(\left|{\sin{\left(x \right)}}\right| \right)} + \int{\tan{\left(x \right)} d x} - 3 {\color{red}{\int{\frac{\sin^{2}{\left(3 x \right)}}{\cos^{2}{\left(3 x \right)}} d x}}}$$

Skriv om i termer av tangens:

$$- \ln{\left(\left|{\sin{\left(x \right)}}\right| \right)} + \int{\tan{\left(x \right)} d x} - 3 {\color{red}{\int{\frac{\sin^{2}{\left(3 x \right)}}{\cos^{2}{\left(3 x \right)}} d x}}} = - \ln{\left(\left|{\sin{\left(x \right)}}\right| \right)} + \int{\tan{\left(x \right)} d x} - 3 {\color{red}{\int{\tan^{2}{\left(3 x \right)} d x}}}$$

Låt $$$u=3 x$$$ vara.

Då $$$du=\left(3 x\right)^{\prime }dx = 3 dx$$$ (stegen kan ses »), och vi har att $$$dx = \frac{du}{3}$$$.

Alltså,

$$- \ln{\left(\left|{\sin{\left(x \right)}}\right| \right)} + \int{\tan{\left(x \right)} d x} - 3 {\color{red}{\int{\tan^{2}{\left(3 x \right)} d x}}} = - \ln{\left(\left|{\sin{\left(x \right)}}\right| \right)} + \int{\tan{\left(x \right)} d x} - 3 {\color{red}{\int{\frac{\tan^{2}{\left(u \right)}}{3} d u}}}$$

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

$$- \ln{\left(\left|{\sin{\left(x \right)}}\right| \right)} + \int{\tan{\left(x \right)} d x} - 3 {\color{red}{\int{\frac{\tan^{2}{\left(u \right)}}{3} d u}}} = - \ln{\left(\left|{\sin{\left(x \right)}}\right| \right)} + \int{\tan{\left(x \right)} d x} - 3 {\color{red}{\left(\frac{\int{\tan^{2}{\left(u \right)} d u}}{3}\right)}}$$

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

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

Alltså,

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

Skriv om och dela upp bråket:

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

Integrera termvis:

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

Tillämpa konstantregeln $$$\int c\, dv = c v$$$ med $$$c=1$$$:

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

Integralen av $$$\frac{1}{v^{2} + 1}$$$ är $$$\int{\frac{1}{v^{2} + 1} d v} = \operatorname{atan}{\left(v \right)}$$$:

$$- v - \ln{\left(\left|{\sin{\left(x \right)}}\right| \right)} + \int{\tan{\left(x \right)} d x} + {\color{red}{\int{\frac{1}{v^{2} + 1} d v}}} = - v - \ln{\left(\left|{\sin{\left(x \right)}}\right| \right)} + \int{\tan{\left(x \right)} d x} + {\color{red}{\operatorname{atan}{\left(v \right)}}}$$

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

$$- \ln{\left(\left|{\sin{\left(x \right)}}\right| \right)} + \int{\tan{\left(x \right)} d x} + \operatorname{atan}{\left({\color{red}{v}} \right)} - {\color{red}{v}} = - \ln{\left(\left|{\sin{\left(x \right)}}\right| \right)} + \int{\tan{\left(x \right)} d x} + \operatorname{atan}{\left({\color{red}{\tan{\left(u \right)}}} \right)} - {\color{red}{\tan{\left(u \right)}}}$$

Kom ihåg att $$$u=3 x$$$:

$$- \ln{\left(\left|{\sin{\left(x \right)}}\right| \right)} + \int{\tan{\left(x \right)} d x} - \tan{\left({\color{red}{u}} \right)} + \operatorname{atan}{\left(\tan{\left({\color{red}{u}} \right)} \right)} = - \ln{\left(\left|{\sin{\left(x \right)}}\right| \right)} + \int{\tan{\left(x \right)} d x} - \tan{\left({\color{red}{\left(3 x\right)}} \right)} + \operatorname{atan}{\left(\tan{\left({\color{red}{\left(3 x\right)}} \right)} \right)}$$

Skriv om tangenten som $$$\tan\left(x\right)=\frac{\sin\left(x\right)}{\cos\left(x\right)}$$$:

$$- \ln{\left(\left|{\sin{\left(x \right)}}\right| \right)} - \tan{\left(3 x \right)} + \operatorname{atan}{\left(\tan{\left(3 x \right)} \right)} + {\color{red}{\int{\tan{\left(x \right)} d x}}} = - \ln{\left(\left|{\sin{\left(x \right)}}\right| \right)} - \tan{\left(3 x \right)} + \operatorname{atan}{\left(\tan{\left(3 x \right)} \right)} + {\color{red}{\int{\frac{\sin{\left(x \right)}}{\cos{\left(x \right)}} d x}}}$$

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

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

Integralen kan omskrivas som

$$- \ln{\left(\left|{\sin{\left(x \right)}}\right| \right)} - \tan{\left(3 x \right)} + \operatorname{atan}{\left(\tan{\left(3 x \right)} \right)} + {\color{red}{\int{\frac{\sin{\left(x \right)}}{\cos{\left(x \right)}} d x}}} = - \ln{\left(\left|{\sin{\left(x \right)}}\right| \right)} - \tan{\left(3 x \right)} + \operatorname{atan}{\left(\tan{\left(3 x \right)} \right)} + {\color{red}{\int{\left(- \frac{1}{u}\right)d u}}}$$

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

$$- \ln{\left(\left|{\sin{\left(x \right)}}\right| \right)} - \tan{\left(3 x \right)} + \operatorname{atan}{\left(\tan{\left(3 x \right)} \right)} + {\color{red}{\int{\left(- \frac{1}{u}\right)d u}}} = - \ln{\left(\left|{\sin{\left(x \right)}}\right| \right)} - \tan{\left(3 x \right)} + \operatorname{atan}{\left(\tan{\left(3 x \right)} \right)} + {\color{red}{\left(- \int{\frac{1}{u} d u}\right)}}$$

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

$$- \ln{\left(\left|{\sin{\left(x \right)}}\right| \right)} - \tan{\left(3 x \right)} + \operatorname{atan}{\left(\tan{\left(3 x \right)} \right)} - {\color{red}{\int{\frac{1}{u} d u}}} = - \ln{\left(\left|{\sin{\left(x \right)}}\right| \right)} - \tan{\left(3 x \right)} + \operatorname{atan}{\left(\tan{\left(3 x \right)} \right)} - {\color{red}{\ln{\left(\left|{u}\right| \right)}}}$$

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

$$- \ln{\left(\left|{\sin{\left(x \right)}}\right| \right)} - \ln{\left(\left|{{\color{red}{u}}}\right| \right)} - \tan{\left(3 x \right)} + \operatorname{atan}{\left(\tan{\left(3 x \right)} \right)} = - \ln{\left(\left|{\sin{\left(x \right)}}\right| \right)} - \ln{\left(\left|{{\color{red}{\cos{\left(x \right)}}}}\right| \right)} - \tan{\left(3 x \right)} + \operatorname{atan}{\left(\tan{\left(3 x \right)} \right)}$$

Alltså,

$$\int{\left(\tan{\left(x \right)} - 3 \tan^{3}{\left(3 x \right)} \cot{\left(3 x \right)} - \cot{\left(x \right)}\right)d x} = - \ln{\left(\left|{\sin{\left(x \right)}}\right| \right)} - \ln{\left(\left|{\cos{\left(x \right)}}\right| \right)} - \tan{\left(3 x \right)} + \operatorname{atan}{\left(\tan{\left(3 x \right)} \right)}$$

Förenkla:

$$\int{\left(\tan{\left(x \right)} - 3 \tan^{3}{\left(3 x \right)} \cot{\left(3 x \right)} - \cot{\left(x \right)}\right)d x} = 3 x - \ln{\left(\left|{\sin{\left(x \right)}}\right| \right)} - \ln{\left(\left|{\cos{\left(x \right)}}\right| \right)} - \tan{\left(3 x \right)}$$

Lägg till integrationskonstanten:

$$\int{\left(\tan{\left(x \right)} - 3 \tan^{3}{\left(3 x \right)} \cot{\left(3 x \right)} - \cot{\left(x \right)}\right)d x} = 3 x - \ln{\left(\left|{\sin{\left(x \right)}}\right| \right)} - \ln{\left(\left|{\cos{\left(x \right)}}\right| \right)} - \tan{\left(3 x \right)}+C$$

$$$\int \left(\tan{\left(x \right)} - 3 \tan^{3}{\left(3 x \right)} \cot{\left(3 x \right)} - \cot{\left(x \right)}\right)\, dx = \left(3 x - \ln\left(\left|{\sin{\left(x \right)}}\right|\right) - \ln\left(\left|{\cos{\left(x \right)}}\right|\right) - \tan{\left(3 x \right)}\right) + C$$$A