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

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Answer 1 · 2018-05-30T19:56:36

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

#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#