CBSE Class 12 Mathematics sample paper 2025-26: question 33(A)

Hints

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Step-by-step solution

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1. Put $x=\tan \theta$, $dx={\sec }^2\theta d\theta$; $x=0\to \theta =0$, $x=1\to \theta =\frac{\pi }{4}$. $I={\int }_0^{\pi /4}\frac{\log (1+\tan \theta )}{1+{\tan }^2\theta }{\sec }^2\theta d\theta$.

   ½ mark

2. $I={\int }_0^{\pi /4}\log (1+\tan \theta )d\theta ={\int }_0^{\pi /4}\log \left[1+\tan \left(\frac{\pi }{4}-\theta \right)\right]d\theta$ [${\int }_0^{a}f(x)dx={\int }_0^{a}f(a-x)dx$].

   1 mark

3. $={\int }_0^{\pi /4}\log \left[1+\frac{1-\tan \theta }{1+\tan \theta }\right]d\theta ={\int }_0^{\pi /4}\log \frac{2}{1+\tan \theta }d\theta$.

   1 mark

4. $={\int }_0^{\pi /4}\log 2d\theta -{\int }_0^{\pi /4}\log (1+\tan \theta )d\theta =\log 2[\theta {]}_0^{\pi /4}-I$.

   1 mark

5. $\Rightarrow 2I=\frac{\pi }{4}\log 2$.

   1 mark

6. $\Rightarrow I=\frac{\pi }{8}\log 2$.

   ½ mark

Answer: $\frac{\pi }{8}\log 2$

Revise the chapter

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