Class 12 Maths · Integrals · Long Answer #64

Question

Evaluate $\int \frac{1}{\cos{\left(a - x \right)} \cos{\left(b - x \right)}}\,dx$. Work on intervals avoiding cosine zeros, with real constants a,b and sin(b−a)≠0.

Accepted Answer

Final Answer: $- \frac{\log{\left(\left|{\frac{\cos{\left(b - x \right)}}{\cos{\left(a - x \right)}}}\right| \right)}}{\sin{\left(a - b \right)}}$ + C Step 1: Use tan(x−a)−tan(x−b)=sin(b−a)/[cos(x−a)cos(x−b)]. Step 2: The integral is [∫tan(x−a)dx−∫tan(x−b)dx]/sin(b−a). Step 3: Integrating tangents gives [−ln|cos(x−a)|+ln|cos(x−b)|]/sin(b−a). Step 4: Therefore the required integral is $- \frac{\log{\left(\left|{\frac{\cos{\left(b - x \right)}}{\cos{\left(a - x \right)}}}\right| \right)}}{\sin{\left(a - b \right)}}$ + C.

Class 12 Maths · Integrals · Long Answer #64

✅ Step-by-Step Solution

📋 CBSE Stepwise Marking Scheme

Master Chapter 7: Integrals

More Practice Questions in Integrals

Related Questions in Integrals