Question

How do I solve this using the substitution method for finding anti derivatives?

Answer 1

A bit blurry, but if I'm making it out correctly, that's

$\displaystyle\int t\sqrt{(t^2-9)^3}\,\mathrm dt=\int t(t^2-9)^{3/2}\,\mathrm dt$

Set $u=t^2-9$. Then $\mathrm du=2t\,\mathrm dt$, or $t\,\mathrm dt=\dfrac{\mathrm du}2$. The integral is then equivalent to

$\displaystyle u^{3/2}\dfrac{\mathrm du}2=\frac12\int u^{3/2}\,\mathrm du$
$=\dfrac12\dfrac{u^{5/2}}{\frac52}+C$
$=\dfrac12\times\dfrac25u^{5/2}+C$
$=\dfrac15u^{5/2}+C$
$=\dfrac15(t^2-9)^{5/2}+C$