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I am suppose to make a substitution and then solve by parts but I can't seem to do anything with it that makes sense.

$\int e^{\cos t}\sin 2t\,dt$

I rewrote as $2e^{\cos t}\sin t\cos t$ and attempted to use u substitution but I end up with

$-2 \int e^{\cos t}{\cos t\,dt}$ and from here I do not know what to do, nothing seems to improve the problem

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    @Jordan: don't forget to upvote/accept answers that you like, and to follow up with comments if you're stuck2012-05-30

2 Answers 2

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HINTS

  1. $\sin(2t) = 2\cos t \sin t$
  2. $\frac{d}{dt} \cos t = - \sin t$
  3. $\displaystyle \int xe^x dx$ is an integral that can be done by parts
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You could use the substitution $u=\cos(t) \implies du = -\sin(t) dt$ and then write the integral as: $2\int{\sin(t)\cos(t)e^{\cos(t)} dt}.$

We then end up with the following expression to evaluate:

$-2\int{ue^{u} du}$

This is trivial to evaluate using integration by parts:

$2\int{e^{u}du-2ue^{u}} = 2e^{u}-2ue^{u}+c_1$

We can now back-substitute to get:

$2e^{\cos(t)}-2\cos(t)e^{\cos(t)}+c_1$

Which can be simplified to give:

$-2e^{\cos(t)}(\cos(t)-1)+c_1$

Hope this helps.