How do you integrate #int 1/(x^3sqrt(9-x^2))dx# using trigonometric substitution?

1 Answer
Noah G
Jan 21, 2017

#-1/54(ln|(3 + sqrt(9 - x^2))/x| + (3sqrt(9 - x^2))/x^2) + C#

Explanation:

Whenever you have an integral with a #√# of the form #sqrt(a^2 - x^2)#, use the substitution #x = asintheta#.

So, let #x = 3sintheta#. Then #dx= 3costhetad theta#.

#=>int 1/((3sintheta)^3sqrt(9 - (3sintheta)^2)) * 3costhetad theta#

#=>int 1/(27sin^3theta sqrt(9 - 9sin^2theta)) * 3costheta d theta#

#=>int 1/(27sin^3thetasqrt(9(1 - sin^2theta))) * 3costheta d theta#

#=>int 1/(27sin^3thetasqrt(9cos^2theta)) * 3costheta d theta#

#=>int 1/(27sin^3theta3costheta) * 3costheta d theta#

#=>int 1/(27sin^3theta) d theta#

#=>1/27int csc^3theta d theta#

This can be integrated as #-(ln|csctheta + cottheta| + cotthetacsctheta)/2 + C#. This is derived using integration by parts. You can see the full proof here.

#=>-1/54(ln|csctheta + cottheta| + cotthetacsctheta) + C#

Now draw a triangle to determine expressions for cosecant and cotangent.

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#=>-1/54(ln|(3 + sqrt(9 - x^2))/x| + (3sqrt(9 - x^2))/x^2) + C#

Hopefully this helps!

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