Question

definite integration from 0 to 2pi [1 / (1+e^sin x) ]

Answer 1

Hope this will help u.
sorry (2pi-x) instead of (2pi+x)

Answer 2

Answer:

π

Step-by-step explanation:

Given,

Integrand: $\int\limits {\frac{1}{1+e^{sinx} } } \, dx$   from 0 to 2pi

= $\int\limits {\frac{1}{1+e^{sinx} } } \, dx +\frac{1}{1+e^{2pi-x} } dx$  from 0 to pi

Since, $\int\limits {f(x)} \, dx$ from 0 to 2a = $\int\limits {f(x)} + f(2a-x)\, dx$ from 0 to a.

= $\int\limit {\frac{1}{1+e^{sinx} } } \, dx + \frac{e^{sinx} }{1+e^{sinx} } dx$ from 0 to pi

= $\int\limits{\frac{1+e^{sinx} }{1+e^{sinx} } } \, dx$ from 0 to pi

= $\int\limits \, dx$ from 0 to pi

= pi.

Hence, the answer is pi.

#SPJ2