Evaluate: `int(sin2x)/((a+bcosx)^2)`

Explanation

Let `I=int(sin2x)/((a+bcosx)^2)dx`
`=>I=int(2sinxcosx)/((a+bcosx)^2)dx`
put `a+bcos x=t`
`=>cos x=(t-a)/b`
`=>-bsinx dx=dt`
`I=-2/b int((t-a)/b)1/t^2*dt`
`=-2/b^2int(t-a)/t^2 dt`
`=-2/b^2[logt-(at^(-2+1))/(-2+1)]+c`
`=-2/b^2[log t+a/t]+c`
`therefore I=-2/b^2[log|a+bcosx|+a/(a+bcosx)]+c`