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(a−b)≠0: log|cos(x+b)/cos(x+a)|/sin(a−b)+C. If a−b=kπ, k∈ℤ: (−1)^k tan(x+a)+C. Exclude the original cosine zeros.
Work on each open interval where the original integrand is real and defined.
Use tan(x+a)−tan(x+b)=sin(a−b)/[cos(x+a)cos(x+b)]. If a−b=kπ, use (−1)ᵏtan(x+a).
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.
Master Chapter 7: Integrals
Practise interactive questions, flashcards, visual mind maps, and timed board tests for Integrals.