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Section 5.4 : More Substitution Rule

13. Evaluate \( \displaystyle \int{{{z^7}{{\left( {8 + 3{z^4}} \right)}^8}\,dz}}\).

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Hint : Use the “obvious” substitution and don’t forget that the substitution can be used more than once and in different ways.
Start Solution

Okay, the “obvious” substitution here is probably,

\[u = 8 + 3{z^4}\hspace{0.25in} \to \hspace{0.25in}du = 12{z^3}dz\hspace{0.25in}\,\,\,\,\,\, \to \hspace{0.25in}\,\,\,\,\,{z^3}dz = \frac{1}{{12}}du\]

however, that doesn’t look like it might work because of the \({z^7}\).

Show Step 2

Let’s do a quick rewrite of the integrand.

\[\int{{{z^7}{{\left( {8 + 3{z^4}} \right)}^8}\,dz}} = \int{{{z^4}{z^3}{{\left( {8 + 3{z^4}} \right)}^8}\,dz}} = \int{{{z^4}{{\left( {8 + 3{z^4}} \right)}^8}\,{z^3}dz}}\] Show Step 3

Now, notice that we can convert all of the \(z\)’s in the integrand except apparently for the \({z^4}\) that is in the front. However, notice from the substitution that we can solve for \({z^4}\) to get,

\[{z^4} = \frac{1}{3}\left( {u - 8} \right)\] Show Step 4

With this we can now do the substitution and evaluate the integral.

\[\begin{align*}\int{{{z^7}{{\left( {8 + 3{z^4}} \right)}^8}\,dz}} & = \frac{1}{{12}}\int{{\frac{1}{3}\left( {u - 8} \right){u^8}du}} = \frac{1}{{36}}\int{{{u^9} - 8{u^8}du}} = \frac{1}{{36}}\left( {\frac{1}{{10}}{u^{10}} - \frac{8}{9}{u^9}} \right) + c\\ & = \require{bbox} \bbox[2pt,border:1px solid black]{{\frac{1}{{36}}\left( {\frac{1}{{10}}{{\left( {8 + 3{z^4}} \right)}^{10}} - \frac{8}{9}{{\left( {8 + 3{z^4}} \right)}^9}} \right) + c}}\end{align*}\]