Integralen av $$$\frac{1}{\cos{\left(a - x \right)} \cos{\left(b - x \right)}}$$$ med avseende på $$$x$$$

Bestäm $$$\int \frac{1}{\cos{\left(a - x \right)} \cos{\left(b - x \right)}}\, dx$$$.

Skriv om integranden:

$${\color{red}{\int{\frac{1}{\cos{\left(a - x \right)} \cos{\left(b - x \right)}} d x}}} = {\color{red}{\int{\left(\frac{\sin{\left(a - x \right)}}{\sin{\left(a - b \right)} \cos{\left(a - x \right)}} - \frac{\sin{\left(b - x \right)}}{\sin{\left(a - b \right)} \cos{\left(b - x \right)}}\right)d x}}}$$

Integrera termvis:

$${\color{red}{\int{\left(\frac{\sin{\left(a - x \right)}}{\sin{\left(a - b \right)} \cos{\left(a - x \right)}} - \frac{\sin{\left(b - x \right)}}{\sin{\left(a - b \right)} \cos{\left(b - x \right)}}\right)d x}}} = {\color{red}{\left(\int{\frac{\sin{\left(a - x \right)}}{\sin{\left(a - b \right)} \cos{\left(a - x \right)}} d x} - \int{\frac{\sin{\left(b - x \right)}}{\sin{\left(a - b \right)} \cos{\left(b - x \right)}} d x}\right)}}$$

Tillämpa konstantfaktorregeln $$$\int c f{\left(x \right)}\, dx = c \int f{\left(x \right)}\, dx$$$ med $$$c=\frac{1}{\sin{\left(a - b \right)}}$$$ och $$$f{\left(x \right)} = \frac{\sin{\left(a - x \right)}}{\cos{\left(a - x \right)}}$$$:

$$- \int{\frac{\sin{\left(b - x \right)}}{\sin{\left(a - b \right)} \cos{\left(b - x \right)}} d x} + {\color{red}{\int{\frac{\sin{\left(a - x \right)}}{\sin{\left(a - b \right)} \cos{\left(a - x \right)}} d x}}} = - \int{\frac{\sin{\left(b - x \right)}}{\sin{\left(a - b \right)} \cos{\left(b - x \right)}} d x} + {\color{red}{\frac{\int{\frac{\sin{\left(a - x \right)}}{\cos{\left(a - x \right)}} d x}}{\sin{\left(a - b \right)}}}}$$

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

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

Alltså,

$$- \int{\frac{\sin{\left(b - x \right)}}{\sin{\left(a - b \right)} \cos{\left(b - x \right)}} d x} + \frac{{\color{red}{\int{\frac{\sin{\left(a - x \right)}}{\cos{\left(a - x \right)}} d x}}}}{\sin{\left(a - b \right)}} = - \int{\frac{\sin{\left(b - x \right)}}{\sin{\left(a - b \right)} \cos{\left(b - x \right)}} d x} + \frac{{\color{red}{\int{\frac{1}{u} d u}}}}{\sin{\left(a - b \right)}}$$

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

$$- \int{\frac{\sin{\left(b - x \right)}}{\sin{\left(a - b \right)} \cos{\left(b - x \right)}} d x} + \frac{{\color{red}{\int{\frac{1}{u} d u}}}}{\sin{\left(a - b \right)}} = - \int{\frac{\sin{\left(b - x \right)}}{\sin{\left(a - b \right)} \cos{\left(b - x \right)}} d x} + \frac{{\color{red}{\ln{\left(\left|{u}\right| \right)}}}}{\sin{\left(a - b \right)}}$$

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

$$\frac{\ln{\left(\left|{{\color{red}{u}}}\right| \right)}}{\sin{\left(a - b \right)}} - \int{\frac{\sin{\left(b - x \right)}}{\sin{\left(a - b \right)} \cos{\left(b - x \right)}} d x} = \frac{\ln{\left(\left|{{\color{red}{\cos{\left(a - x \right)}}}}\right| \right)}}{\sin{\left(a - b \right)}} - \int{\frac{\sin{\left(b - x \right)}}{\sin{\left(a - b \right)} \cos{\left(b - x \right)}} d x}$$

Tillämpa konstantfaktorregeln $$$\int c f{\left(x \right)}\, dx = c \int f{\left(x \right)}\, dx$$$ med $$$c=\frac{1}{\sin{\left(a - b \right)}}$$$ och $$$f{\left(x \right)} = \frac{\sin{\left(b - x \right)}}{\cos{\left(b - x \right)}}$$$:

$$\frac{\ln{\left(\left|{\cos{\left(a - x \right)}}\right| \right)}}{\sin{\left(a - b \right)}} - {\color{red}{\int{\frac{\sin{\left(b - x \right)}}{\sin{\left(a - b \right)} \cos{\left(b - x \right)}} d x}}} = \frac{\ln{\left(\left|{\cos{\left(a - x \right)}}\right| \right)}}{\sin{\left(a - b \right)}} - {\color{red}{\frac{\int{\frac{\sin{\left(b - x \right)}}{\cos{\left(b - x \right)}} d x}}{\sin{\left(a - b \right)}}}}$$

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

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

Integralen blir

$$\frac{\ln{\left(\left|{\cos{\left(a - x \right)}}\right| \right)}}{\sin{\left(a - b \right)}} - \frac{{\color{red}{\int{\frac{\sin{\left(b - x \right)}}{\cos{\left(b - x \right)}} d x}}}}{\sin{\left(a - b \right)}} = \frac{\ln{\left(\left|{\cos{\left(a - x \right)}}\right| \right)}}{\sin{\left(a - b \right)}} - \frac{{\color{red}{\int{\frac{1}{u} d u}}}}{\sin{\left(a - b \right)}}$$

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

$$\frac{\ln{\left(\left|{\cos{\left(a - x \right)}}\right| \right)}}{\sin{\left(a - b \right)}} - \frac{{\color{red}{\int{\frac{1}{u} d u}}}}{\sin{\left(a - b \right)}} = \frac{\ln{\left(\left|{\cos{\left(a - x \right)}}\right| \right)}}{\sin{\left(a - b \right)}} - \frac{{\color{red}{\ln{\left(\left|{u}\right| \right)}}}}{\sin{\left(a - b \right)}}$$

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

$$\frac{\ln{\left(\left|{\cos{\left(a - x \right)}}\right| \right)}}{\sin{\left(a - b \right)}} - \frac{\ln{\left(\left|{{\color{red}{u}}}\right| \right)}}{\sin{\left(a - b \right)}} = \frac{\ln{\left(\left|{\cos{\left(a - x \right)}}\right| \right)}}{\sin{\left(a - b \right)}} - \frac{\ln{\left(\left|{{\color{red}{\cos{\left(b - x \right)}}}}\right| \right)}}{\sin{\left(a - b \right)}}$$

Alltså,

$$\int{\frac{1}{\cos{\left(a - x \right)} \cos{\left(b - x \right)}} d x} = \frac{\ln{\left(\left|{\cos{\left(a - x \right)}}\right| \right)}}{\sin{\left(a - b \right)}} - \frac{\ln{\left(\left|{\cos{\left(b - x \right)}}\right| \right)}}{\sin{\left(a - b \right)}}$$

Förenkla:

$$\int{\frac{1}{\cos{\left(a - x \right)} \cos{\left(b - x \right)}} d x} = \frac{\ln{\left(\left|{\cos{\left(a - x \right)}}\right| \right)} - \ln{\left(\left|{\cos{\left(b - x \right)}}\right| \right)}}{\sin{\left(a - b \right)}}$$

Lägg till integrationskonstanten:

$$\int{\frac{1}{\cos{\left(a - x \right)} \cos{\left(b - x \right)}} d x} = \frac{\ln{\left(\left|{\cos{\left(a - x \right)}}\right| \right)} - \ln{\left(\left|{\cos{\left(b - x \right)}}\right| \right)}}{\sin{\left(a - b \right)}}+C$$

$$$\int \frac{1}{\cos{\left(a - x \right)} \cos{\left(b - x \right)}}\, dx = \frac{\ln\left(\left|{\cos{\left(a - x \right)}}\right|\right) - \ln\left(\left|{\cos{\left(b - x \right)}}\right|\right)}{\sin{\left(a - b \right)}} + C$$$A