ज्ञात करें ` int (2sin 2 phi - cos phi)/(6-cos^2phi -4 sinphi)d phi`

Explanation

`I= int (2sin 2 phi - cos phi)/(6-cos^2phi -4 sinphi)d phi`
`=int ((4 sin phi -1)cosphi)/(6-(1-sin^2 phi)-4sinphi)d phi = int ((4sin phi-1)cosphi)/(sin^2 phi -4sin phi +5)dphi`
`sin phi= t` रखे, तो `cos phi d phi=dt," ":." "I=int (4t-1)/(t^2-t+5) dt `
माना कि `(4t - 1) = lamda ""d/(dt)(t^2 -4t +5) + mu`
` rArr " "(4t -1) = lamda ( 2t -4) +mu`
t के समान घाट के गुणांकों को बराबर करने पर हमे मिलता है,
`2lamda = 4, -4lamda + mu = -1 4" "rArr" "lamda = 2, mu=7`
`:." "I=int (lamda(2t-4))/(t^2 - 4t +5)dt`
`2int ((2t-4))/(t^2 - 4t +5)dt +7 int 1/(t^2 -4t +1^2) dt`
`=2log | t^2-4t +5| +7 int 1/((t-2)^2+1^2)dt`
`=2log | t^2 -4t+5|+7 tan^(-1)(t-2)+c`
अतः `I=2 log| sin^2 phi - sin phi + 5 | + 7 tan^(-1) ( sin phi -2) +c`