Question

How do you integrate `int 1/(x^4sqrt(x^2+3))` by trigonometric substitution?

Answer 1

`sqrt(x^2+3)/(9x)-(x^2+3)^(3/2)/(27x^3)+C`

Explanation:

`int dx/(x^4*sqrt(x^2+3))`

After using `x=sqrt3*tany` and `dx=sqrt3*(secy)^2*dy` transforms, I found

`int (sqrt3*(secy)^2*dy)/((sqrt3*tany)^4*sqrt((sqrt3*tany)^2+3))`

=`int (sqrt3*(secy)^2*dy)/((9(tany)^4*sqrt((sqrt3*secy)^2))`

=`int (sqrt3*(secy)^2*dy)/((9(tany)^4*sqrt3*secy)`

=`1/9int (secy*dy)/(tany)^4`

=`1/9int (coty)^4*secy*dy`

=`1/9int (cosy/siny)^4*(dy/cosy)`

=`1/9int ((cosy)^3*dy)/(siny)^4`

=`1/9int ((cosy)^2*cosy*dy)/(siny)^4`

=`1/9int ((1-(siny)^2)*cosy*dy)/(siny)^4`

After using `z=siny` and `dz=cosy*dy` transforms, it became

`1/9int ((1-z^2)*dz)/z^4`

=`1/9int z^(-4)*dz-1/9int z^(-2)*dz`

=`1/9z^(-1)-1/27z^(-3)+C`

=`1/9(siny)^(-1)-1/27(siny)^(-3)+C`

=`1/9cscy-1/27(cscy)^3+C`

After using `x=sqrt3*tany`, `tany=x/sqrt3`, `secy=sqrt(x^2+3)/sqrt3` and `cscy=secy/tany=sqrt(x^2+3)/x` inverse transforms, I found

`sqrt(x^2+3)/(9x)-(x^2+3)^(3/2)/(27x^3)+C`