Evaluate: `int(sinx)/(sin4x)dx`

Explanation

`I=int(sinx)/(sin4x)dx=int(sinx)/(2sin2x cos2x)dx`
`=int(sinx)/(4sinx cos x cos2x)dx`
`=(1)/(4)int(1)/(cosx cos2x)dx=(1)/(4)int(cosx)/(cos^(2)x cos2x)dx`
`=(1)/(4)int(cosx)/((1-sin^(2)x)(1-2sin^(2)x))dx`
Putting `sinx =t` and `cosx dx=dt,` we get
`I=(1)/(4)int(dt)/((1-t^(2))(1-2t^(2)))`
`=(1)/(4)int[(2)/(1-2t^(2))-(1)/(1-t^(2))]dt`
`= -(1)/(4)int(1)/(1-t^(2))dt+(2)/(4)int(1)/(1-(sqrt(2)t)^(2))dt`
`= -(1)/(4)xx(1)/(2)log|(1+t)/(1-t)|+(1)/(2).(1)/(2sqrt(2))log|(1+sqrt(2)t)/(1-sqrt(2)t)| +C`
`= -(1)/(8)log|(1+sinx)/(1-sinx)|+(1)/(4sqrt(2))log|(1+sqrt(2)sinx)/(1-sqrt(2)sinx)| +C`