Calculus II - Trig Substitutions

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Section 7.3 : Trig Substitutions

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7. Use a trig substitution to eliminate the root in ${\left( {{x^2} - 8x + 21} \right)^{\frac{3}{2}}}$ .

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We know that in order to do a trig substitution we really need a sum or difference of a term with a variable squared and a number. This clearly does not fit into that form. However, that doesn’t mean that we can’t do some algebraic manipulation on the quantity under the root to get into a form that we can do a trig substitution on.

Because the quantity under the root is a quadratic polynomial we know that we can complete the square on it to turn it into something like what we need for a trig substitution.

Here is the completing the square work.

$\begin{align*}{x^2} - 8x + 21 & = {x^2} - 8x + 16 - 16 + 21\hspace{0.75in}{\left[ {\frac{1}{2}\left( { - 8} \right)} \right]^2} = {\left[ { - 4} \right]^2} = 16\\ & = {\left( {x - 4} \right)^2} + 5\end{align*}$

So, after completing the square the term can be written as,

${\left( {{x^2} - 8x + 21} \right)^{\frac{3}{2}}} = {\left( {{{\left( {x - 4} \right)}^2} + 5} \right)^{\frac{3}{2}}} = {\left[ {\sqrt {{{\left( {x - 4} \right)}^2} + 5} } \right]^3}$

Note that we also explicitly put the root into the problem as well.

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So, in this case we see that we have a sum of a squared term with a variable in it and a number. This suggests that tangent is the correct trig function to use for the substation.

Now, to get the coefficient on the trig function notice that we need to turn the 1 (i.e. the coefficient of the squared term) into a 5 once we’ve done the substitution. With that in mind it looks like the substitution should be,

$x - 4 = \sqrt 5 \tan \left( \theta \right)$

Now, all we have to do is actually perform the substitution and eliminate the root.

Show Step 3

$\begin{align*}{\left( {{x^2} - 8x + 21} \right)^{\frac{3}{2}}} & = {\left[ {\sqrt {{{\left( {x - 4} \right)}^2} + 5} } \right]^3} = {\left[ {\sqrt {{{\left( {\sqrt 5 \tan \left( \theta \right)} \right)}^2} + 5} } \right]^3}\\ & = {\left[ {\sqrt {5{{\tan }^2}\left( \theta \right) + 5} } \right]^3} = {\left[ {\sqrt 5 \sqrt {{{\tan }^2}\left( \theta \right) + 1} } \right]^3}\\ & = {\left[ {\sqrt 5 \sqrt {{{\sec }^2}\left( \theta \right)} } \right]^3} = \require{bbox} \bbox[2pt,border:1px solid black]{{{5^{\frac{3}{2}}}{{\left| {\sec \left( \theta \right)} \right|}^3}}}\end{align*}$

Note that because we don’t know the values of $\theta$ we can’t determine if the secant is positive or negative and so cannot get rid of the absolute value bars here.