`int_(0)^(pi)(xdx)/(1+sinx)`

Explanation

माना `I=int_(0)^(pi)(x)/(1+sinx)dx" …(1)"`
`rArr" "I=int_(0)^(pi)(pi-x)/(1+sin(pi-x))dx" "` [ प्रगुण 4 से ]
`" "=int_(0)^(pi)(pi-x)/(1+sinx)dx" ...(2)"`
समीकरण (1 ) और (2 ) को जोड़ने पर ,
`2I=int_(0)^(pi)(x+pi-x)/(1+sinx)dx`
`=pi int_(0)^(pi)(1)/(1+sinx).(1-sinx)/(1-sinx)dx`
`=pi int_(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 tanx)dx`
`=pi[tanx-secx]_(0)^(pi)`
`=pi[(tanpi-secpi)-(tan 0 - sec 0)]`
`=pi(1+1)=2pi`
`rArr" "I=pi`