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int_(0)^(pi)(x dx)/(1+sin x) |
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Answer» Solution :`"Let I"= int_(0)^(pi) (x)/(1+sinx)DX` `RARR ""I=int_(0)^(pi) (pi-x)/(1+sin (pi-x))dx` `=int_(0)^(pi)(pi-x)/(1+sin x)dx` Adding equations (1) and (2) `2I =int_(0)^(pi)(x+pi-x)/(1+sin x)dx` `=piint_(0)^(pi)(1)/(1+sinx).(1-sin x)/(1-sin x)dx` `=piint_(0)^(pi)(1-sinx)/(1-sin^(2)x)dx` `=pi int_(0)^(pi)(1-sinx)/(COS^(2)x)dx` `=pi int_(0)^(pi)(sec^(2)x-secx tan x) dx` `=pi [tan x-sec x]_(0)^(pi)` `=pi [(tan pi-sec pi)-(tan 0-sec 0)]` ` =pi (1+1) =2pi` `rArr ""I=pi` |
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