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  • Please solve RD Sharma Class 12 Chapter 19 Definite Integrals Exercise 19.4 (a) Question 1 maths textbook solution.

Please solve RD Sharma Class 12 Chapter 19 Definite Integrals Exercise 19.4 (a) Question 1 maths textbook solution.

Answers (1)

Answer : \pi

Given : \int_{0}^{2 \pi} \frac{e^{\sin x}}{e^{\sin x}+e^{-\sin x}} d x

Hint : Use the formula of \int_{0}^{2 \pi} f(x) d x=\int_{0}^{2 \pi} f(2 \pi-x) d x

Solution : I=\int_{0}^{2 \pi} \frac{e^{\sin x}}{e^{\sin x}+e^{-\sin x}} d x-------(1)

\begin{aligned} &\int_{0}^{2 \pi} \frac{e^{\sin (2 \pi-x)}}{e^{\sin (2 \pi-x)}+e^{-\sin (2 \pi-x)}} d x \\ &\int_{0}^{2 \pi} f(x) d x=\int_{0}^{2 \pi} f(2 \pi-x) d x \end{aligned}

I=\int_{0}^{2 \pi} \frac{e^{-\sin x}}{e^{-\sin x}+e^{\sin x}} d x \quad[\sin (2 \pi-x)=-\sin x]---------(2)

Add (1) and (2)

\begin{aligned} &2 I=\int_{0}^{2 \pi} \frac{e^{-\sin x}}{e^{-\sin x}+e^{\sin x}} d x+\int_{0}^{2 \pi} \frac{e^{\sin x}}{e^{-\sin x}+e^{\sin x}} d x \\ &2 I=\int_{0}^{2 \pi} \frac{e^{-\sin x}+e^{\sin x}}{e^{-\sin x}+e^{\sin x}} d x \\ &2 I=\int_{0}^{2 \pi} d x \end{aligned}

2I=2\pi

I=\pi

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