Calculus II - Trig Substitutions

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Section 7.3 : Trig Substitutions

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3. Use a trig substitution to eliminate the root in \({\left( {7{t^2} - 3} \right)^{\frac{5}{2}}}\).

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Start Solution

First, notice that there really is a root here as the term can be written as,

\[{\left( {7{t^2} - 3} \right)^{\frac{5}{2}}} = {\left[ {{{\left( {7{t^2} - 3} \right)}^{\frac{1}{2}}}} \right]^5} = {\left[ {\sqrt {7{t^2} - 3} } \right]^5}\]

Now, we need to figure out which trig function to use for the substitution. To determine this notice that (ignoring the numbers) the quantity under the root looks similar to the identity,

\[{\sec ^2}\left( \theta \right) - 1 = {\tan ^2}\left( \theta \right)\]

So, it looks like secant is probably the correct trig function to use for the substitution. Now, we need to deal with the numbers on the two terms.

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To get the coefficient on the trig function notice that we need to turn the 7 into a 3 once we’ve substituted the trig function in for \(t\) and squared the substitution out. With that in mind it looks like the substitution should be,

\[t = \frac{{\sqrt 3 }}{{\sqrt 7 }}\sec \left( \theta \right)\]

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

Show Step 3

\[\begin{align*}{\left( {7{t^2} - 3} \right)^{\frac{5}{2}}} & = {\left[ {\sqrt {7{t^2} - 3} } \right]^5}\\ & = {\left[ {\sqrt {7{{\left( {\frac{{\sqrt 3 }}{{\sqrt 7 }}\sec \left( \theta \right)} \right)}^2} - 3} } \right]^5} = {\left[ {\sqrt {7\left( {\frac{3}{7}} \right){{\sec }^2}\left( \theta \right) - 3} } \right]^5}\\ & = {\left[ {\sqrt {3{{\sec }^2}\left( \theta \right) - 3} } \right]^5} = {\left[ {\sqrt 3 \sqrt {{{\sec }^2}\left( \theta \right) - 1} } \right]^5}\\ & = {\left[ {\sqrt 3 \sqrt {{{\tan }^2}\left( \theta \right)} } \right]^5} = \require{bbox} \bbox[2pt,border:1px solid black]{{{3^{\frac{5}{2}}}{{\left| {\tan \left( \theta \right)} \right|}^5}}}\end{align*}\]

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