Section 7.1 : Integration by Parts
7. Evaluate \( \displaystyle \int{{{t^7}\sin \left( {2{t^4}} \right)\,dt}}\) .
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The first step here is to pick \(u\) and \(dv\) and, in this case, we’ll need to be careful how we chose them.
If we follow the model of many of the examples/practice problems to this point it is tempting to let \(u\) be \({t^7}\) and to let \(dv\) be \(\sin \left( {2{t^4}} \right)\) .
However, this will lead to some real problems. To compute \(v\) we’d have to integrate the sine and because of the \({t^4}\) in the argument this is not possible. In order to integrate the sine we would have to have a \({t^3}\) in the integrand as well in order to a substitution as shown below,
\[\int{{{t^3}\sin \left( {2{t^4}} \right)\,dt}} = \frac{1}{8}\int{{\sin \left( w \right)\,dw}} = - \frac{1}{8}\cos \left( {2{t^4}} \right) + c\hspace{0.25in}w = 2{t^4}\]Now, this may seem like a problem, but in fact it’s not a problem for this particular integral. Notice that we actually have 7 \(t\)’s in the integral and there is no reason that we can’t split them up as follows,
\[ \int{{{t^7}\sin \left( {2{t^4}} \right)\,dt}} = \int{{{t^4}\,{t^3}\sin \left( {2{t^4}} \right)\,dt}}\]After doing this we can now choose \(u\) and \(dv\) as follows,
\[u = {t^4}\hspace{0.5in}dv = {t^3}\sin \left( {2{t^4}} \right)\,dt\] Show Step 2Next, we need to compute \(du\) (by differentiating \(u\)) and \(v\) (by integrating \(dv\)).
\[\begin{align*}u & = {t^4} & \hspace{0.5in} & \to & \hspace{0.25in}du & = 4{t^3}dt\\ dv & = \,{t^3}\sin \left( {2{t^4}} \right)\,dt & \hspace{0.5in} & \to & \hspace{0.25in}v & = - \frac{1}{8}\cos \left( {2{t^4}} \right)\end{align*}\] Show Step 3Plugging \(u\), \(du\), \(v\) and \(dv\) into the Integration by Parts formula gives,
\[ \int{{{t^7}\sin \left( {2{t^4}} \right)\,dt}} = - \frac{1}{8}{t^4}\cos \left( {2{t^4}} \right) + \frac{1}{2}\int{{{t^3}\cos \left( {2{t^4}} \right)\,dt}}\] Show Step 4At this point, notice that the new integral just requires the same Calculus I substitution that we used to find \(v\). So, all we need to do is evaluate the new integral and we’ll be done.
\[ \int{{{t^7}\sin \left( {2{t^4}} \right)\,dt}} = \require{bbox} \bbox[2pt,border:1px solid black]{{ - \frac{1}{8}{t^4}\cos \left( {2{t^4}} \right) + \frac{1}{{16}}\sin \left( {2{t^4}} \right) + c}}\]Do not get so locked into patterns for these problems that you end up turning the patterns into “rules” on how certain kinds of problems work. Most of the easily seen patterns are also easily broken (as this problem has shown).
Because we (as instructors) tend to work a lot of “easy” problems initially they also tend to conform to the patterns that can be easily seen. This tends to lead students to the idea that the patterns will always work and then when they run into one where the pattern doesn’t work they get in trouble. So, be careful!
Note as well that we’re not saying that patterns don’t exist and that it isn’t useful to recognize them. You just need to be careful and understand that there will, on occasion, be problems where it will look like a pattern you recognize, but in fact will not quite fit the pattern and another approach will be needed to work the problem.
Show Alternate SolutionNote that there is an alternate solution to this problem. We could use the substitution \(w = 2{t^4}\) as the first step as follows.
\[w = 2{t^4}\hspace{0.5in} \to \hspace{0.5in}dw = 8{t^3}dt \hspace{0.5in} \& \hspace{0.5in} {t^4} = \frac{1}{2}w\] \[ \int{{{t^7}\sin \left( {2{t^4}} \right)\,dt}} = \int{{{t^4}\,{t^3}\sin \left( {2{t^4}} \right)\,dt}} = \int{{\left( {\frac{1}{2}w} \right)\left( {\frac{1}{8}} \right)\sin \left( w \right)\,dw}} = \int{{\frac{1}{{16}}w\sin \left( w \right)\,dw}}\]We won’t avoid integration by parts as we can see here, but it is somewhat easier to see it this time. Here is the rest of the work for this problem.
\[\begin{align*}u & = \frac{1}{{16}}w & \hspace{0.5in} & \to & \hspace{0.25in}du & = \frac{1}{{16}}dw\\ dv & = \,\sin \left( w \right)\,dw & \hspace{0.5in} & \to & \hspace{0.25in}v & = - \cos \left( w \right)\end{align*}\] \[\int{{{t^7}\sin \left( {2{t^4}} \right)\,dt}} = - \frac{1}{{16}}w\cos \left( w \right) + \frac{1}{{16}}\int{{\cos \left( w \right)\,dw}} = - \frac{1}{{16}}w\cos \left( w \right) + \frac{1}{{16}}\sin \left( w \right) + c\]As the final step we just need to substitution back in for \(w\).
\[\int{{{t^7}\sin \left( {2{t^4}} \right)\,dt}} = - \frac{1}{8}{t^4}\cos \left( {2{t^4}} \right) + \frac{1}{{16}}\sin \left( {2{t^4}} \right) + c\]