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Section 12.7 : Calculus with Vector Functions

7. Evaluate $$\displaystyle \int{{\vec r\left( t \right)\,dt}}$$, where $$\displaystyle \vec r\left( t \right) = {t^3}\,\vec i - \frac{{2t}}{{{t^2} + 1}}\vec j + {\cos ^2}\left( {3t} \right)\vec k$$.

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There really isn’t a lot to do here with this problem. All we need to do is integrate of all the components of the vector.

\begin{align*}\int{{\vec r\left( t \right)\,dt}} & = \int{{{t^3}\,dt\,}}\vec i - \int{{\frac{{2t}}{{{t^2} + 1}}\,dt\,}}\vec j + \int{{{{\cos }^2}\left( {3t} \right)\,dt}}\,\vec k\\ & = \int{{{t^3}\,dt\,}}\vec i - \int{{\frac{{2t}}{{{t^2} + 1}}\,dt\,}}\vec j + \int{{\frac{1}{2}\left( {1 + \cos \left( {6t} \right)} \right)\,dt}}\,\vec k\\ & = \require{bbox} \bbox[2pt,border:1px solid black]{{\frac{1}{4}{t^4}\,\vec i - \ln \left| {{t^2} + 1} \right|\,\vec j + \frac{1}{2}\left( {t + \frac{1}{6}\sin \left( {6t} \right)} \right)\vec k + \vec c}}\end{align*}

Don’t forget the basic integration rules such as the substitution rule (second term) and some of the basic trig formulas (half angle and double angle formulas) you need to do some of the integrals (third term).

We didn’t put a lot of the integration details into the solution on the assumption that you do know your integration skills well enough to follow what is going on. If you are somewhat rusty with your integration skills then you’ll need to go back to the integration material from both Calculus I and Calculus II to refresh your integration skills.