CBSE Class 12 Mathematics 2026: question 27(a)

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. Split the integral: $I=\int \frac{x}{1-\cos x}dx-\int \frac{\sin x}{1-\cos x}dx$

   ½ mark

2. Use $1-\cos x=2{\sin }^2\frac{x}{2}$ and $\sin x=2\sin \frac{x}{2}\cos \frac{x}{2}$:

   $I=\frac{1}{2}\int x{\text{cosec}}^2\frac{x}{2}dx-\int \cot \frac{x}{2}dx$

   1 mark

3. Integrate the first term by parts, taking $x$ as the first function and $\int {\text{cosec}}^2\frac{x}{2}dx=-2\cot \frac{x}{2}$:

   $I=\frac{1}{2}\left[-2x\cot \frac{x}{2}-\int (-2\cot \frac{x}{2})dx\right]-\int \cot \frac{x}{2}dx=-x\cot \frac{x}{2}+\int \cot \frac{x}{2}dx-\int \cot \frac{x}{2}dx$

   1 mark

4. The two remaining integrals cancel, so $I=-x\cot \frac{x}{2}+C$.

   ½ mark

Answer: $-x\cot \frac{x}{2}+C$

Revise the chapter

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