The value of

Question:

The value of $\int_{-\pi / 2}^{\pi / 2} \frac{1}{1+e^{\sin x}} d x$ is

  1. $\pi$

  2. $\frac{3 \pi}{2}$

  3. $\frac{\pi}{4}$

  4. $\frac{\pi}{2}$


Correct Option: , 4

Solution:

$I=\int_{-\pi / 2}^{\pi / 2} \frac{1}{1+e^{\sin x}} d x$ $\ldots(1)$

Apply King property

$I=\int_{-z / 2}^{\pi / 2} \frac{1}{1+e^{-\sin x}} d x=\int_{-\pi / 2}^{\pi / 2} \frac{e^{\sin x}}{1+e^{\sin x} x} d x \ldots(2)$

Add (1) & (2)

$2 \mathbf{I}=\int_{-\pi / 2}^{\pi / 2} \mathrm{~d} \mathrm{x}=\pi$

$I=\frac{\pi}{2}$

 

Leave a comment

Close

Click here to get exam-ready with eSaral

For making your preparation journey smoother of JEE, NEET and Class 8 to 10, grab our app now.

Download Now