Question

Trig Substitution?

∫(√x^2-81)/(x^4)dx

I can't seem to figure out this problem. Please help.

Answer 1

`1/243(x^2-81)^(3/2)/x^3+C`

Explanation:

`int(sqrt(x^2-81))/x^4dx`

we shall be using the identity

`1+tan^2u=sec^2`

`x=9secu=>dx=9secutanu`

now substitute

`intsqrt(81sec^4u-81)/(9^4sec^4u)xx9secutanudu`

simplify,

NOTE: some steps are omitted for the reader to verify themselves.

`int(sqrt(81(sec^2u-1))/(9^4sec^4u))xx9secutanudu`

`=81/9^4int(tanu/sec^4u)xxsecutanudu`

`=1/81int(tan^2u/sec^3u)du`

change to sines and cosines

`=1/81int(sin^2u/cos^2u)xxcos^3udu`

`=1/81intsin^2ucosudu`

`=1/81xx1/3sin^3u+C`

we now need to substitute back for `u`

`secu=x/9=>cosu=9/x, " & " sinu=sqrt(x^2-81)`

we have

`=1/243(sqrt(x^2-81)^3/(x^3))+C`

`=1/243(x^2-81)^(3/2)/x^3+C`