Answer:
[tex]\dfrac{\ln | \sec (x-b)- \ln | \sec (x-a)}{\sin (a-b)}+\text{C}[/tex]
Step-by-step explanation:
Fundamental Theorem of Calculus
[tex]\displaystyle \int \text{f}(x)\:\text{d}x=\text{F}(x)+\text{C} \iff \text{f}(x)=\dfrac{\text{d}}{\text{d}x}(\text{F}(x))[/tex]
If differentiating takes you from one function to another, then integrating the second function will take you back to the first with a constant of integration.
Given indefinite integral:
[tex]\displaystyle \int \sec(x-a) \sec (x-b)\:\text{d}x[/tex]
[tex]\boxed{\begin{minipage}{3.5 cm}\underline{Trigonometric Identity}\\\\$\sec \theta=\dfrac{1}{\cos \theta}$\\\end{minipage}}[/tex]
Therefore:
[tex]\implies \displaystyle \int \dfrac{1}{\cos(x-a)} \cdot \dfrac{1}{\cos (x-b)}\:\text{d}x[/tex]
[tex]\implies \displaystyle \int \dfrac{1}{\cos(x-a)\cos (x-b)} \:\text{d}x[/tex]
[tex]\textsf{Multiply the integral by }\dfrac{\sin (a-b)}{\sin (a-b)}:[/tex]
[tex]\implies \displaystyle \int \dfrac{1}{\cos(x-a)\cos (x-b)} \cdot \dfrac{\sin (a-b)}{\sin (a-b)}\:\text{d}x[/tex]
Take the constant outside the integral:
[tex]\implies \displaystyle \dfrac{1}{\sin (a-b)}\int \dfrac{\sin (a-b)}{\cos(x-a)\cos (x-b)} \:\text{d}x[/tex]
Rewrite the numerator:
[tex]\implies \displaystyle \dfrac{1}{\sin (a-b)}\int \dfrac{\sin [(x-b)-(x-a)]}{\cos(x-a)\cos (x-b)} \:\text{d}x[/tex]
[tex]\boxed{\begin{minipage}{6 cm}\underline{Trigonometric Identity}\\\\$\sin (A \pm B)=\sin A \cos B \pm \cos A \sin B$\\\end{minipage}}[/tex]
Therefore:
[tex]\implies \displaystyle \dfrac{1}{\sin (a-b)}\int \dfrac{\sin(x-b) \cos (x-a)-\cos (x-b) \sin (x-a)}{\cos(x-a)\cos (x-b)} \:\text{d}x[/tex]
[tex]\implies \displaystyle \dfrac{1}{\sin (a-b)}\int \dfrac{\sin(x-b) \cos (x-a)}{\cos(x-a)\cos (x-b)} -\dfrac{\cos (x-b) \sin (x-a)}{{\cos(x-a)\cos (x-b)}}\:\text{d}x[/tex]
[tex]\implies \displaystyle \dfrac{1}{\sin (a-b)}\int \dfrac{\sin(x-b)}{\cos (x-b)} -\dfrac{\sin (x-a)}{\cos(x-a)}\:\text{d}x[/tex]
[tex]\boxed{\begin{minipage}{3.5 cm}\underline{Trigonometric Identity}\\\\$\tan \theta=\dfrac{\sin \theta}{\cos \theta}$\\\end{minipage}}[/tex]
Therefore:
[tex]\implies \displaystyle \dfrac{1}{\sin (a-b)}\int \tan(x-b)-\tan(x-a)\:\text{d}x[/tex]
[tex]\boxed{\begin{minipage}{4.3 cm}\underline{Integrating $\tan x$}\\\\$\displaystyle \int \tan x\:\text{d}x=\ln | \sec x|+\text{C}$\end{minipage}}[/tex]
Therefore:
[tex]\implies \dfrac{1}{\sin (a-b)}\left[ \ln | \sec (x-b)- \ln | \sec (x-a)\right]+\text{C}[/tex]
[tex]\implies \dfrac{\ln | \sec (x-b)- \ln | \sec (x-a)}{\sin (a-b)}+\text{C}[/tex]
Learn more about indefinite integrals here:
https://brainly.com/question/28308973
