Integralen av $$$\cos{\left(\theta \right)} \cot{\left(\theta \right)}$$$

Bestäm $$$\int \cos{\left(\theta \right)} \cot{\left(\theta \right)}\, d\theta$$$.

Skriv om integranden:

$${\color{red}{\int{\cos{\left(\theta \right)} \cot{\left(\theta \right)} d \theta}}} = {\color{red}{\int{\frac{\cos^{2}{\left(\theta \right)}}{\sin{\left(\theta \right)}} d \theta}}}$$

Multiplicera täljaren och nämnaren med en sinus och skriv allt annat i termer av cosinus, med hjälp av formeln $$$\sin^2\left(\alpha \right)=-\cos^2\left(\alpha \right)+1$$$ med $$$\alpha=\theta$$$:

$${\color{red}{\int{\frac{\cos^{2}{\left(\theta \right)}}{\sin{\left(\theta \right)}} d \theta}}} = {\color{red}{\int{\frac{\sin{\left(\theta \right)} \cos^{2}{\left(\theta \right)}}{1 - \cos^{2}{\left(\theta \right)}} d \theta}}}$$

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

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

Alltså,

$${\color{red}{\int{\frac{\sin{\left(\theta \right)} \cos^{2}{\left(\theta \right)}}{1 - \cos^{2}{\left(\theta \right)}} d \theta}}} = {\color{red}{\int{\left(- \frac{u^{2}}{1 - u^{2}}\right)d u}}}$$

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

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

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

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

Integrera termvis:

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

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

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

Utför partialbråksuppdelning (stegen kan ses »):

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

Integrera termvis:

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

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

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

Låt $$$v=u + 1$$$ vara.

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

Alltså,

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

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

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

Kom ihåg att $$$v=u + 1$$$:

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

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

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

Låt $$$v=u - 1$$$ vara.

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

Alltså,

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

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

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

Kom ihåg att $$$v=u - 1$$$:

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

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

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

Alltså,

$$\int{\cos{\left(\theta \right)} \cot{\left(\theta \right)} d \theta} = \frac{\ln{\left(\left|{\cos{\left(\theta \right)} - 1}\right| \right)}}{2} - \frac{\ln{\left(\left|{\cos{\left(\theta \right)} + 1}\right| \right)}}{2} + \cos{\left(\theta \right)}$$

Lägg till integrationskonstanten:

$$\int{\cos{\left(\theta \right)} \cot{\left(\theta \right)} d \theta} = \frac{\ln{\left(\left|{\cos{\left(\theta \right)} - 1}\right| \right)}}{2} - \frac{\ln{\left(\left|{\cos{\left(\theta \right)} + 1}\right| \right)}}{2} + \cos{\left(\theta \right)}+C$$

$$$\int \cos{\left(\theta \right)} \cot{\left(\theta \right)}\, d\theta = \left(\frac{\ln\left(\left|{\cos{\left(\theta \right)} - 1}\right|\right)}{2} - \frac{\ln\left(\left|{\cos{\left(\theta \right)} + 1}\right|\right)}{2} + \cos{\left(\theta \right)}\right) + C$$$A