Evaluate the following integral: ∫(e^(sin x))/(e^(sin x) + e^(-sin x)) dx, x ∈ [0, 2π]

Question

Evaluate the following integral:

\(\int\limits_{0}^{2\pi}\cfrac{e^{sin\,\text x}}{e^{sin\,\text x}+e^{-sin\,\text x}}d\text x \)

Answer 1

let us assume I =  \(\int\limits_{0}^{2\pi}\cfrac{e^{sin\,\text x}}{e^{sin\,\text x}+e^{-sin\,\text x}}d\text x \).... equation 1

By property, we know that \(\int\limits_{a}^{b} \)f(x) dx = \(\int\limits_{a}^{b} \)f(a + b - x)dx

I = \(\int\limits_{0}^{2\pi}\cfrac{e^{sin(2\pi-\text x)}}{e^{sin(2\pi-\text x)}+e^{-sin(2\pi-\text x)}}d\text x \).....equation 2

Adding equation 1 and 2

2I = \(\int\limits_{0}^{2\pi}\cfrac{e^{sin\,\text x}}{e^{sin\,\text x}+e^{-sin\,\text x}}d\text x \) + \(\int\limits_{0}^{2\pi}\cfrac{e^{sin(2\pi-\text x)}}{e^{sin(2\pi-\text x)}+e^{-sin(2\pi-\text x)}}d\text x \)

We know sin(2π - x) = -sin x

Thus

2I = \(\int\limits_{0}^{2\pi}\cfrac{e^{sin\,\text x}}{e^{sin\,\text x}+e^{-sin\,\text x}}d\text x \) + \(\int\limits_{0}^{2\pi}\cfrac{e^{-sin\,\text x}}{e^{-sin\,\text x}+e^{sin\,\text x}}d\text x \)

We know