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Class 12 Integrals · Long Answer3 Marksindefinite_integration

Class 12 Maths · Integrals · Long Answer #84

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

✅ Step-by-Step Solution

Final Answer: \(\frac{\log{\left(\left|{\frac{\sin{\left(a - x \right)}}{\cos{\left(b - x \right)}}}\right| \right)}}{\cos{\left(a - b \right)}}\) + C
  1. Use cot(x−a)+tan(x−b)=cos(a−b)/[sin(x−a)cos(x−b)].
  2. Divide by cos(a−b) and integrate the cotangent and tangent separately.
  3. This gives [ln|sin(x−a)|−ln|cos(x−b)|]/cos(a−b).
  4. Therefore the required integral is \(\frac{\log{\left(\left|{\frac{\sin{\left(a - x \right)}}{\cos{\left(b - x \right)}}}\right| \right)}}{\cos{\left(a - b \right)}}\) + C.

📋 CBSE Stepwise Marking Scheme

  • [0.5 marks] Use cot(x−a)+tan(x−b)=cos(a−b)/[sin(x−a)cos(x−b)].
  • [1 mark] Divide by cos(a−b) and integrate the cotangent and tangent separately.
  • [1 mark] This gives [ln|sin(x−a)|−ln|cos(x−b)|]/cos(a−b).
  • [0.5 marks] Therefore the required integral is \(\frac{\log{\left(\left|{\frac{\sin{\left(a - x \right)}}{\cos{\left(b - x \right)}}}\right| \right)}}{\cos{\left(a - b \right)}}\) + C.

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