# Finding the integral of $e^{\cos t} \sin 2t$

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: press here "show steps" – Ilya May 30 '12 at 12:32
@Jordan: don't forget to upvote/accept answers that you like, and to follow up with comments if you're stuck – mixedmath May 30 '12 at 12:54

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.

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