1 Answers
`I=int (1)/(sinx-sin2x)dx`
`=int(1)/((sinx-2sinx cos x))dx`
`=int (1)/(sinx(1-2cosx))dx`
`=int (sin x)/(sin^(2)x(1-2cosx))dx`
`=int (sin x)/((1-cos^(2)x)(1-2cosx))dx`
Putting `cos x=t` and `-sinx dx =dt`
or `sinx dx= -dt`, we get
`I=int(-dt)/((1-t^(2))(1-2t))`
`=int(1)/((t-1)(1+t)(1-2t))dt`
`=int((1)/((t-1)(2)(-1))+(1)/((-2)(1+t)(3))+(1)/((-1//2)(3//2)(1-2t)))dt`
`= -(1)/(2)log|1-t|-(1)/(6)log|1+t|+(2)/(3)log|1-2t|+C`
`= -(1)/(2)log|1-cos x|-(1)/(6)log|1+cosx|+(2)/(3)log|1-2cos x|+C`
4 views
Answered