# The workings of Trig Substitution

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?

• 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? – Kitter Catter Sep 17 at 21:35
• 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. – Burt Sep 17 at 21:38
• RHS is right hand side LHS is left hand side – Kitter Catter Sep 17 at 21:39
• How does that help me in this case? – Burt Sep 17 at 21:47
• You should be able to find the antiderivative of $\sec^3\theta$. – Andrew Chin Sep 17 at 23:37

• @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. – BallBoy Sep 18 at 19:37