Answer of this question

Hi Nitin

The Textual version of the answer is

Integration (0 to pi) ecosx / ecosx + e-cosx 

Evaluate: ∫_0^π▒〖e^cos⁡x /(e^cos⁡x   +  e^〖-cos〗⁡x  ) dx〗 

Let   I=∫_0^π▒〖e^cos⁡x /(e^cos⁡x   +  e^〖-cos〗⁡x  ) dx〗 "  "

  I= ∫_0^π▒〖e^cos⁡〖(π - x)〗 /(e^cos⁡〖(π - x)〗   +  e^〖-cos〗⁡〖(π - x)〗  ) dx〗 " "

   I=∫_0^π▒〖e^〖-cos〗⁡x /(e^〖-cos〗⁡x   +  e^(〖-(-cos〗⁡x)) ) dx〗

   I=∫_0^π▒〖e^〖-cos〗⁡x /(e^〖-cos〗⁡x   +  e^cos⁡x  ) dx〗

Adding (1) and  (2) i.e. (1) + (2)

  I+I=∫_0^π▒〖e^cos⁡x /(e^cos⁡x   +  e^〖-cos〗⁡x  ) dx〗 + ∫_0^π▒〖e^〖-cos〗⁡x /(e^〖-cos〗⁡x   +  e^cos⁡x  ) dx〗

   2I=∫_0^π▒〖(e^cos⁡x   +  e^〖-cos〗⁡x )/(e^cos⁡x   +  e^〖-cos〗⁡x  ) dx〗

   2I  =∫_0^π▒〖1 .〗⁡dx

    2I=[x]_0^π

    2I  =π-0

     2I =π

     I=π/2   

Notes:

Using P4 : ∫_0^a▒〖f(x)dx=〗 ∫_0^a▒f(a-x)dx

(As cos (π-x)=-cos⁡x)

Written on Jan. 26, 2017, 12:13 p.m.