CBSE Class 12 Mathematics 2025: question 32(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. Let $I={\int }_0^{\pi }\frac{x\tan x}{\sec x+\tan x}dx$ …(i). Using ${\int }_0^{a}f(x)dx={\int }_0^{a}f(a-x)dx$ with $\tan (\pi -x)=-\tan x$ and $\sec (\pi -x)=-\sec x$: $I={\int }_0^{\pi }\frac{(\pi -x)\tan x}{\sec x+\tan x}dx$ …(ii).

   1½ marks

2. Adding (i) and (ii): $2I=\pi {\int }_0^{\pi }\frac{\tan x}{\sec x+\tan x}dx=\pi {\int }_0^{\pi }\frac{\tan x(\sec x-\tan x)}{{\sec }^{2}x-{\tan }^{2}x}dx$

   1 mark

3. Since ${\sec }^{2}x-{\tan }^{2}x=1$: $2I=\pi {\int }_0^{\pi }(\sec x\tan x-{\tan }^{2}x)dx=\pi {\int }_0^{\pi }(\sec x\tan x-{\sec }^{2}x+1)dx$

   1 mark

4. $2I=\pi {\left[\sec x-\tan x+x\right]}_0^{\pi }$

   1 mark

5. $2I=\pi [(-1-0+\pi )-(1-0+0)]=\pi (\pi -2)$, so $I=\frac{\pi }{2}(\pi -2)$.

   ½ mark

Answer: $I=\frac{\pi }{2}(\pi -2)$

Revise the chapter

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