Section 5.4 : More Substitution Rule
2. Evaluate \( \displaystyle \int{{7{x^3}\cos \left( {2 + {x^4}} \right) - 8{x^3}{{\bf{e}}^{2 + {x^{\,4}}}}\,dx}}\).
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Don’t get too excited about the fact that there are two terms in this integrand. Each term requires the same substitution,
\[u = 2 + {x^4}\]so we’ll simply use that in both terms.
If you aren’t comfortable with the basic substitution mechanics you should work some problems in the previous section as we’ll not be putting in as much detail with regards to the basics in this section. The problems in this section are intended for those that are fairly comfortable with the basic mechanics of substitutions and will involve some more “advanced” substitutions.
Show Step 2Here is the differential work for the substitution.
\[du = 4{x^3}\,dx\hspace{0.25in} \to \hspace{0.25in}\,{x^3}dx = \frac{1}{4}du\]Before doing the actual substitution it might be convenient to factor an \({x^3}\) out of the integrand as follows.
\[\int{{7{x^3}\cos \left( {2 + {x^4}} \right) - 8{x^3}{{\bf{e}}^{2 + {x^{\,4}}}}\,dx}} = \int{{\left[ {7\cos \left( {2 + {x^4}} \right) - 8{{\bf{e}}^{2 + {x^{\,4}}}}} \right]{x^3}dx}}\]Doing this should make the differential part (i.e. the \(du\) part) of the substitution clearer.
Now, doing the substitution and evaluating the integral gives,
\[\begin{align*}\int{{7{x^3}\cos \left( {2 + {x^4}} \right) - 8{x^3}{{\bf{e}}^{2 + {x^{\,4}}}}\,dx}} &= \frac{1}{4}\int{{7\cos \left( u \right) - 8{{\bf{e}}^u}du}}\\ & = \frac{1}{4}\left[ {7\sin \left( u \right) - 8{{\bf{e}}^u}} \right] + c = \require{bbox} \bbox[2pt,border:1px solid black]{{\frac{1}{4}\left[ {7\sin \left( {2 + {x^4}} \right) - 8{{\bf{e}}^{2 + {x^{\,4}}}}} \right] + c}}\end{align*}\]Be careful when dealing with the \(dx\) substitution here. Make sure that the \(\frac{1}{4}\) gets multiplied times the whole integrand and not just one of the terms. You can do this either by using parenthesis around the whole integrand or pulling the \(\frac{1}{4}\) completely out of the integral (as we’ve done here).
Do not forget to go back to the original variable after evaluating the integral!