Calculus II - Integration by Parts

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Section 7.1 : Integration by Parts

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3. Evaluate $\displaystyle \int{{\left( {3t + {t^2}} \right)\sin \left( {2t} \right)\,dt}}$ .

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The first step here is to pick $u$ and $dv$ . We want to choose $u$ and $dv$ so that when we compute $du$ and $v$ and plugging everything into the Integration by Parts formula the new integral we get is one that we can do or will at least be an integral that will be easier to deal with.

With that in mind it looks like the following choices for $u$ and $dv$ should work for us.

$u = 3t + {t^2}\hspace{0.5in}dv = \sin \left( {2t} \right)\,dt$ Show Step 2

Next, we need to compute $du$ (by differentiating $u$ ) and $v$ (by integrating $dv$ ).

$\begin{align*}u & = 3t + {t^2} & \hspace{0.5in} & \to & \hspace{0.25in}du & = \left( {3 + 2t} \right)dt\\ dv & = \sin \left( {2t} \right)\,dt & \hspace{0.5in} & \to & \hspace{0.25in}v & = - \frac{1}{2}\cos \left( {2t} \right)\end{align*}$ Show Step 3

Plugging $u$ , $du$ , $v$ and $dv$ into the Integration by Parts formula gives,

$\int{{\left( {3t + {t^2}} \right)\sin \left( {2t} \right)\,dt}} = - \frac{1}{2}\left( {3t + {t^2}} \right)\cos \left( {2t} \right) + \frac{1}{2}\int{{\left( {3 + 2t} \right)\cos \left( {2t} \right)\,dt}}$ Show Step 4

Now, the new integral is still not one that we can do with only Calculus I techniques. However, it is one that we can do another integration by parts on and because the power on the $t$ ’s have gone down by one we are heading in the right direction.

So, here are the choices for $u$ and $dv$ for the new integral.

$\begin{align*}u & = 3 + 2t & \hspace{0.5in} & \to & \hspace{0.25in}du & = 2dt\\ dv & = \cos \left( {2t} \right)\,dt & \hspace{0.5in} & \to & \hspace{0.25in}v & = \frac{1}{2}\sin \left( {2t} \right)\end{align*}$ Show Step 5

Okay, all we need to do now is plug these new choices of $u$ and $dv$ into the new integral we got in Step 3 and finish the problem out.

$\begin{align*}\int{{\left( {3t + {t^2}} \right)\sin \left( {2t} \right)\,dt}} & = - \frac{1}{2}\left( {3t + {t^2}} \right)\cos \left( {2t} \right) + \frac{1}{2}\left[ {\frac{1}{2}\left( {3 + 2t} \right)\sin \left( {2t} \right) - \int{{\sin \left( {2t} \right)\,dt}}} \right]\\ & = - \frac{1}{2}\left( {3t + {t^2}} \right)\cos \left( {2t} \right) + \frac{1}{2}\left[ {\frac{1}{2}\left( {3 + 2t} \right)\sin \left( {2t} \right) + \frac{1}{2}\cos \left( {2t} \right)} \right] + c\\ & = \require{bbox} \bbox[2pt,border:1px solid black]{{ - \frac{1}{2}\left( {3t + {t^2}} \right)\cos \left( {2t} \right) + \frac{1}{4}\left( {3 + 2t} \right)\sin \left( {2t} \right) + \frac{1}{4}\cos \left( {2t} \right) + c}}\end{align*}$