Evaluate: `int1/(sqrt(sin^3xsin(x+alpha)))\ dx ,\ alpha!=npi,\ \ n in Z`

Explanation

`int1/(sqrt(sin^3xsin(x+alpha)))\ dx`
`sin^3xsin(x+alpha)`
`=sin^4x[cos a + cot x sin a] `
so we can write integration as ,
`int 1/sin^2x xx1/sqrt(cos (alpha)+cot x sin(alpha)) dx`
let `cos a + cot x sin alpha = t`
` - cosec^2x sin alpha dx= dt `
`cosec^2x dx= 1/ sin alpha dt `
Now put in the equation
`=-1/sin alpha int(t)^(-1/2) dt`
`=-1/sin alpha [2sqrt t +c]`
put the value and you will get
`=-2/sin alpha sqrt(cos alpha+cot x*sin alpha)+c`
`=>sin^3x sin(x+alpha)=sin^4x(cos alpha+cot x*sin alpha)`
`=>(sin^3x sin(x+alpha))/(sin^4x)=cos alpha+cot x*sin alpha`
`=>(sin(x+alpha))/sinx= cos alpha+cot x*sin alpha`
thus
answer`=-2/sinx sqrt((sin(x+alpha))/sin x)+c`