CBSE Class 12 Mathematics 2023: question 31(b)

Hints

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Step-by-step solution

Each step shows the marks it earns, as an examiner would award them. Checked by an independent re-solve from the question paper.

1. Write $\sin (a-b)=\sin [(x-b)-(x-a)]$. Then

   $I=\frac{1}{\sin (a-b)}\int \frac{\sin [(x-b)-(x-a)]}{\cos (x-a)\cos (x-b)}dx$

   1 mark

2. Expanding the sine:

   $I=\frac{1}{\sin (a-b)}\int \left[\frac{\sin (x-b)\cos (x-a)}{\cos (x-a)\cos (x-b)}-\frac{\cos (x-b)\sin (x-a)}{\cos (x-a)\cos (x-b)}\right]dx$

   ½ mark

3. $I=\frac{1}{\sin (a-b)}\int [\tan (x-b)-\tan (x-a)]dx$

   ½ mark

4. $I=\frac{1}{\sin (a-b)}\left[\log |\sec (x-b)|-\log |\sec (x-a)|\right]+C=\frac{1}{\sin (a-b)}\log \left|\frac{\cos (x-a)}{\cos (x-b)}\right|+C$

   1 mark

Answer: $\frac{1}{\sin (a-b)}\log \left|\frac{\cos (x-a)}{\cos (x-b)}\right|+C$

Revise the chapter

• Chapter: Integrals (notes, formulas, where marks are lost and every past question)