Use a Trig Substitution to eliminate the root in $(x^2-8x+21)^\frac32$

This is the work for this problem:

  • complete the square: $$x^2-8x+21 +16-16$$$$((x-4)^2+5)^\frac32$$

  • Use the substitution $x-4=\sqrt5\tan\theta$ $$\sqrt{(\sqrt5\tan\theta)^2+5)^3}$$ $$=\sqrt{(5(\tan^2\theta+1))^3}$$ $$=[\sqrt5\sqrt{\sec^2\theta}]^3$$ $$=5^\frac32|\sec^3\theta|$$

I believe all this is correct, but how does $(x^2-8x+21)^\frac32 = 5^\frac32|\sec^3\theta|$?

Shouldn't there be another step of resubstituting my $x-4=\sqrt{\tan\theta}$? How am I supposed to do that - my last step had no tangents remaining?

  • 1
    $\begingroup$ Presumably this is in the context of doing an integral? Maybe integrating the RHS is a bit more tractable/apparent on how to solve than integrating the LHS? $\endgroup$ – Kitter Catter Sep 17 at 21:35
  • $\begingroup$ what is RHS and LHS? and yes - in the context of integrating although this specific problem was not an interval. The general question applies to integrals too. $\endgroup$ – Burt Sep 17 at 21:38
  • $\begingroup$ RHS is right hand side LHS is left hand side $\endgroup$ – Kitter Catter Sep 17 at 21:39
  • $\begingroup$ How does that help me in this case? $\endgroup$ – Burt Sep 17 at 21:47
  • $\begingroup$ You should be able to find the antiderivative of $\sec^3\theta$. $\endgroup$ – Andrew Chin Sep 17 at 23:37

You can resubstitute, but you'll end up with the same thing you started with.

The idea of trig substitution is that by performing a change of variables, we can change to a variable in which integration is possible, then change back after integrating. It has the same purpose as the regular substitutions you've probably already learnt to do. As in those substitutions, resubstituting without integrating defeats the purpose and would just undo what you've done.

  • $\begingroup$ So in integration resubstitution would be a necessary step? How do you figure out the substitutions? $\endgroup$ – Burt Sep 18 at 19:18
  • $\begingroup$ @burt In integration you resubstitute so that your final answer is expressed in terms of the same variable as the initial, which is often desired. How you figure out the substitutions is a different question -- you're looking for a substitution which will make integration possible, and you recognize those after a lot of practice. $\endgroup$ – BallBoy Sep 18 at 19:23
  • $\begingroup$ Right, that makes sense. But, how do I know how to resubstitute - the answer gotten after integration usually contains a different trig function than the original substitution. $\endgroup$ – Burt Sep 18 at 19:33
  • $\begingroup$ @burt You can use trig identities to relate different trig functions. I usually draw a right triangle -- so in the case the triangle would have adjacent side $\sqrt5$, opposite side $x-4$, and you can get the hypotenuse from the Pythagorean theorem, then you can figure out all the trig functions. $\endgroup$ – BallBoy Sep 18 at 19:37

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