Evaluate `int(dx)/((sinx-sin2x)).`

Explanation

`int(dx)/((sinx-sin2x))=int (dx)/((sin s-2 sin x cos x))`
`=int (dx)/(sinx(1-2cos x ))=int(sinx)/(sin^(2)x(1-2 cos x))dx`
`intj (sin x)/((1-cos^(2) x)(1-2cos x))dx`
`=-int(dt)/((1-t^(2))(1-2t)),`where cos x=t
`= int (dt)/((t-1)(t+1)(1-2t)).`
`Let (1)/((t-1)(t+1)(1-2t))=(A)/((t-1))+(B)/((t+1))+( C)/((1-2t)).`
`then , 1-=A(t+1)(1-2t)+B(t-1)(1-2t)+C(t-1)(t+1).`
Putting `t=1` in (i) , we get `A=(-1)/(2)`
Putting `t=-1` in (ii), we get `B=(-1)/(6).`
Putting `t=(1)/(2)` in (ii), we get `C=(-4)/(3).`
`therefore I=-(1)/(2)int(dt )/((t-1))-(1)/(6). int (dt) /((t+1))-(4)/(3).int (dt)/((1-2t))`
`=-(1)/(2)log|t-1|-(1)/(6)log |t+1|+(2)/(3).int(-2dt)/((1-2t))`
`=-(1)/(2) log |t-1|-(1)/(6)log |t+1|+(2)/(3)log|1-2t+C`
`=-(1)/(2) log |cos x-1|-(1)/(6)log |cos x+1|+(2)/(3)log |1-2 cos x|+c.`