How do you integrate $1 / (t^3(t^2 - 9)^(1/2))$ ?

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Answer 1 · 2015-05-05T16:38:24

Well, this is a bit hard !

Let's $u = sqrt(t^2-9)$
$du = t/sqrt(t^2-9)dt$

$dt = sqrt(t^2-9)/tdu$

Integral become :

$=>intsqrt(t^2-9)/(t^4sqrt(t^2-9))du = int1/t^4du$

$1/t^4 = 1/(u^2+9)^2$

So now we have $=> int 1/(u^2+9)^2du$

can't do partial fraction so let's $u = 3tan(v)$

$=>du = 3/cos^2(v)dv$

$=>1/(u^2+9)^2 = 1/(9tan^2(v)+9)^2 = cos^4(v)/81$

So now we have

$=>intcos^4(v)/81*3/cos^2(v)dv= 3/81intcos^2(v)dv$

$=>3/162int1+cos(2v)dv = 1/54int1+cos(2v)dv$

$=> 1/54[v+1/2sin(2v)]+C$

Substitute back for $v = arctan(1/3u)$

$=>1/54[arctan(1/3u)+1/2sin(2arctan(1/3u))]$

Substitute again for $u =sqrt(t^2-9)$

$=>1/54[arctan(1/3sqrt(t^2-9))+1/2sin(2arctan(1/3sqrt(t^2-9)))]$

How do you integrate 1 / (t^3(t^2 - 9)^(1/2))1 / (t^3(t^2 - 9)^(1/2))?