Evaluate: `int(dx)/(sinx(3+cos^2x)`

Answer

`(1)/(4)log|(y-1)/(y+1)|-(1)/(4sqrt(3))"tan"^(-1)(y)/(sqrt(3))+C`

Explanation

`int(dx)/(sinx(3+cos^(2)x))=int(sinxdx)/(sin^(2)x(3+cos^(2)x))`
`=int(sinxdx)/((1-cos^(2)x)(3+cos^(2)x))`
`=int(dy)/((y^(2)-1)(y^(2)+3)) " " ("Putting " cosx=y)`
`=(1)/(4)int[(1)/(y^(2)-1)-(1)/(y^(2)+3)]dy`
`=(1)/(4)log|(y-1)/(y+1)|-(1)/(4sqrt(3))"tan"^(-1)(y)/(sqrt(3))+C`