`(1)/(sqrt(sin^(3)x.sin(x+alpha)))`

Explanation

`"Let I"=int (dx)/(sqrt(sin^(3)x.sin(x+alpha)))`
`=int (dx)/(sqrt(sin^(3)x.(sinx cosalpha+cos x sinalpha)))`
`=int (cosec^(2)xdx)/(sqrt(cos alpha+cot.x sin alpha))`
Put `cos alpha+cot x.sin alpha=t`
`therefore -cosec^(2)x.sinalphadx=dt`
`therefore cosec^(2)x dx=-(dt)/(sin alpha)`
`therefore I=int (1)/(sqrtt)((-dt)/(sin alpha))`
`=(-1)/(sinalpha)int t^(-(1)/(2))dt`
`=(-1)/(sin alpha).(t^((1)/(2)))/(((1)/(2)))+c`
`=(-2)/(sin x)sqrtt+c`
`=-2 cosec alpha sqrt(cos alpha+cot x sin alpha)+c`