`int(sin2x)/(a+bcosx)^2dx`

Answer

` -(2)/(b^(2))[log|a+b cosx|+(a)/(a+b cosx)]+C`

Explanation

Let ` I=int(sin2x)/(a+bcosx)^(2)dx=int(2sinx cosx)/(a+bcosx)^(2)dx.`
` " Putting " a+bcosx=t " and "-bsinxdx=dt, " we get " `
` I=-(2)/(b)int(1)/(t^(2))((t-a)/(b))dt [ :' a+bcosx=t, cosx=(t-a)/(b)] `
`=-(2)/(b^(2))int((1)/(t)-(a)/(t^(2)))dt=-(2)/(b^(2))[log|t|+(a)/(t)]+C`
`= -(2)/(b^(2))[log|a+b cosx|+(a)/(a+b cosx)]+C`