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### Section 5.4 : More Substitution Rule

9. Evaluate $$\displaystyle \int{{7\left( {3y + 2} \right){{\left( {4y + 3{y^2}} \right)}^3} + \sin \left( {3 + 8y} \right)\,dy}}$$.

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Hint : You can only do one substitution per integral. At this point we should know how to “break” integrals up so that we can get the terms that require different substitutions into different integrals.
Start Solution

Clearly each term needs a separate substitution. So, we’ll first need to split up the integral as follows.

$\int{{7\left( {3y + 2} \right){{\left( {4y + 3{y^2}} \right)}^3} + \sin \left( {3 + 8y} \right)\,dy}} = \int{{7\left( {3y + 2} \right){{\left( {4y + 3{y^2}} \right)}^3}\,dy}} + \int{{\sin \left( {3 + 8y} \right)\,dy}}$ Show Step 2

The substitutions needed for each integral are then,

$u = 4y + 3{y^2}\hspace{0.75in}v = 3 + 8y$

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 each substitution.

\begin{align*}du & = \left( {4 + 6y} \right)dy = 2\left( {3y + 2} \right)dy\hspace{0.25in}\,\,\,\, \to \hspace{0.25in}\,\left( {3y + 2} \right)dy = \frac{1}{2}du\\ dv & = 8\,dy\hspace{0.5in} \to \hspace{0.25in}\,dy = \frac{1}{8}dv\end{align*}

Now, doing the substitutions and evaluating the integrals gives,

\begin{align*}\int{{7\left( {3y + 2} \right){{\left( {4y + 3{y^2}} \right)}^3} + \sin \left( {3 + 8y} \right)\,dy}} & = \frac{7}{2}\int{{{u^3}du}} + \frac{1}{8}\int{{\sin \left( v \right)\,dv}} = \frac{7}{8}{u^4} - \frac{1}{8}\cos \left( v \right) + c\\ & = \require{bbox} \bbox[2pt,border:1px solid black]{{\frac{7}{8}{{\left( {4y + 3{y^2}} \right)}^4} - \frac{1}{8}\cos \left( {3 + 8y} \right) + c}}\end{align*}

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