use-the-table-of-integrals-on-reference-pages-6-10-to-evaluate-the-integral-displaystyle-int-et-sin-

Question

Use the Table of Integrals on Reference Pages 6-10 to evaluate the integral.
 
$ \displaystyle \int e^t \sin (\alpha t - 3)\ dt $

Foster Wisusik · Accepted Answer

Okay, so this question wants us to evaluate this integral. So to do that, we want to put it in the form of one of our integral Sze in the table. So we would like to make this sign of Alfa Times something so well, it&#x27;s factor out an Alfa from our sign. And now let&#x27;s call that Are you? So you is equal to T minus three over Alfa. So that means that d U is equal to D T and T is equal to you plus three over Alfa. So now we can do some back substituting and get the integral of e to the T, which is you. Plus three over Alfa Times. Sign of Alfa Times, you do you. And then as a last simplification, we can pull this e to the three Alfa outside of are integral sign due to exponentially ation loss. And now we&#x27;re left with this, which is exactly one of our forms in our integral table. So if we go over there, we find this formula and in this case, are a is just equal the Alfa and we have ah e to the three over Alfa outside. Still Okay, so now we gotta back substitute times. Well, you he to the u So eat of the U is T minus river Alfa over one plus Alfa squared. And that is all times Sign of our original argument minus Alfa Co sign of our original argument, plus C. And now we can see one final simplification. So this e to the three over Alfa cancels with this E to the minus through Alfa. So then we get our final answer off E to the T over one plus Alfa squared times Sign of Alfa T minus three minus alfa Co sign of Alfa T minus three plus C And this is our final form of our answer.

Problem

Use the Table of Integrals on Reference Pages 6-1…

Problem 30 Hard Difficulty

Use the Table of Integrals on Reference Pages 6-10 to evaluate the integral.

$ \displaystyle \int e^t \sin (\alpha t - 3)\ dt $

Answer

$\frac{e^{t}}{1+\alpha^{2}}(\sin (\alpha t-3)-\alpha \cos (\alpha t-3))+C$

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