CBSE Class 12 Mathematics 2023: question 28(a)

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. Let $I={\int }_0^{2\pi }\frac{1}{1+e^{\sin x}}dx$.

   Using ${\int }_0^{a}f(x)dx={\int }_0^{a}f(a-x)dx$:

   $I={\int }_0^{2\pi }\frac{1}{1+e^{\sin (2\pi -x)}}dx$

   1 mark

2. Since $\sin (2\pi -x)=-\sin x$:

   $I={\int }_0^{2\pi }\frac{1}{1+e^{-\sin x}}dx={\int }_0^{2\pi }\frac{e^{\sin x}}{e^{\sin x}+1}dx$

   1 mark

3. Adding the two forms of $I$:

   $2I={\int }_0^{2\pi }\frac{1+e^{\sin x}}{1+e^{\sin x}}dx={\int }_0^{2\pi }1dx=2\pi$

   ½ mark

4. $I=\pi$

   ½ mark

Answer: $I=\pi$

Revise the chapter

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