Evaluate \(\int \frac{1}{\cos{\left(a - x \right)} \cos{\left(b - x \right)}}\, dx\).
💡 Concept Hints
Identify the domain and the inner and outer functions, or the relevant one-sided limits.
Apply the appropriate differentiation or limit rule, keep all chain factors, and check exceptional points.
✅ Step-by-Step Solution
Final Answer: For sin(b−a)≠0: log|cos(b−x)/cos(a−x)|/sin(b−a)+C. If b−a=kπ, k∈ℤ: (−1)^k tan(x−a)+C. Exclude the original cosine zeros.
Work where the original cosines and sin(b−a) are nonzero.
Use tan(x−a)−tan(x−b)=sin(b−a)/[cos(x−a)cos(x−b)]. Integrate the tangent difference.
An antiderivative is F(x)=\(- \frac{\log{\left(\left|{\frac{\cos{\left(b - x \right)}}{\cos{\left(a - x \right)}}}\right| \right)}}{\sin{\left(a - b \right)}}\).
Differentiate this expression to check that F′(x)=\(\frac{1}{\cos{\left(a - x \right)} \cos{\left(b - x \right)}}\). Add an arbitrary constant C on each domain component.
If b−a=kπ, k∈Z, the integral is (−1)ᵏtan(x−a)+C instead.
Master Chapter 7: Integrals
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