CBSE Class 12 Mathematics 2026: question 26

Hints

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Step-by-step solution

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1. Integrate by parts taking ${\tan }^{-1}x$ as first function:
   ${\int }_0^{1}x{\tan }^{-1}xdx={\left[{\tan }^{-1}x\cdot \frac{x^2}{2}\right]}_0^1-{\int }_0^1\frac{1}{1+x^2}\cdot \frac{x^2}{2}dx$

   1 mark

2. $=\frac{\pi }{8}-\frac{1}{2}{\int }_0^1\frac{(1+x^2)-1}{1+x^2}dx=\frac{\pi }{8}-\frac{1}{2}{\left[x-{\tan }^{-1}x\right]}_0^1$

   1 mark

3. $=\frac{\pi }{8}-\frac{1}{2}\left(1-\frac{\pi }{4}\right)=\frac{\pi }{8}-\frac{1}{2}+\frac{\pi }{8}=\frac{\pi }{4}-\frac{1}{2}$

   1 mark

Answer: $\frac{\pi }{4}-\frac{1}{2}$

Revise the chapter

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