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### Section 5.4 : More Substitution Rule

6. Evaluate $$\displaystyle \int{{20{{\bf{e}}^{2 - 8w}}\sqrt {1 + {{\bf{e}}^{2 - 8w}}} \, + 7{w^3} - 6\,\,\sqrt[3]{w}\,dw}}$$.

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Hint : Recall that terms that do not need substitutions should not be in the integral when the substitution is being done. At this point we should know how to “break” integrals up so that we can get the terms that require a substitution into a one integral and those that don’t into another integral.
Start Solution

Clearly the first term needs a substitution while the second and third terms don’t. So, we’ll first need to split up the integral as follows.

$\int{{20{{\bf{e}}^{2 - 8w}}\sqrt {1 + {{\bf{e}}^{2 - 8w}}} \, + 7{w^3} - 6\,\,\sqrt[3]{w}\,dw}} = \int{{20{{\bf{e}}^{2 - 8w}}\sqrt {1 + {{\bf{e}}^{2 - 8w}}} \,dw}} + \int{{7{w^3} - 6\,\,\sqrt[3]{w}\,dw}}$ Show Step 2

The substitution needed for the first integral is then,

$u = 1 + {{\bf{e}}^{2 - 8w}}$

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 3

Here is the differential work for the substitution.

$du = - 8{{\bf{e}}^{2 - 8w}}dw\hspace{0.25in} \to \hspace{0.25in}\,{{\bf{e}}^{2 - 8w}}dw = - \frac{1}{8}du$

Now, doing the substitutions and evaluating the integrals gives,

\begin{align*}\int{{20{{\bf{e}}^{2 - 8w}}\sqrt {1 + {{\bf{e}}^{2 - 8w}}} \, + 7{w^3} - 6\,\,\sqrt[3]{w}\,dw}} & = - \frac{{20}}{8}\int{{{u^{\frac{1}{2}}}\,du}} + \int{{7{w^3} - 6{w^{\frac{1}{3}}}\,dw}}\\ & = - \frac{5}{3}{u^{\frac{3}{2}}} + \frac{7}{4}{w^4} - \frac{9}{2}{w^{\frac{4}{3}}} + c\\ & = \require{bbox} \bbox[2pt,border:1px solid black]{{ - \frac{5}{3}{{\left( {1 + {{\bf{e}}^{2 - 8w}}} \right)}^{\frac{3}{2}}} + \frac{7}{4}{w^4} - \frac{9}{2}{w^{\frac{4}{3}}} + c}}\end{align*}

Do not forget to go back to the original variable after evaluating the integral!