CBSE Class 12 Mathematics 2023: question 26

Hints

Open one hint at a time and try again before the next.

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. $I={\int }_0^{\pi /2}\log \frac{\sin x}{2\cos x}dx={\int }_0^{\pi /2}\log \left(\frac{\tan x}{2}\right)dx$ ... (1)

   ½ mark

2. Using ${\int }_0^{a}f(x)dx={\int }_0^{a}f(a-x)dx$, and $\tan \left(\frac{\pi }{2}-x\right)=\cot x$:

   $I={\int }_0^{\pi /2}\log \left(\frac{\cot x}{2}\right)dx$ ... (2)

   1 mark

3. Adding (1) and (2):

   $2I={\int }_0^{\pi /2}\log \left(\frac{\tan x}{2}\cdot \frac{\cot x}{2}\right)dx={\int }_0^{\pi /2}\log \frac{1}{4}dx$

   ½ mark

4. $2I=\log \frac{1}{4}[x{]}_0^{\pi /2}=\frac{\pi }{2}\log \frac{1}{4}$

   ½ mark

5. $I=\frac{\pi }{4}\log \frac{1}{4}=-\frac{\pi }{2}\log 2$

   ½ mark

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

Revise the chapter

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