1. Can't evaluate this integral

$\displaystyle \int_0^{\pi}e^{cost}sin(2t)dt$

Work I have:
$\displaystyle Let u = e^{cost}$

$\displaystyle du=(-e^{cost}sint)dt$

But I don't know what to do from here since it is a sin(2t) in the original integral.

2. Originally Posted by Em Yeu Anh
$\displaystyle \int_0^{\pi}e^{cost}sin(2t)dt$

Work I have:
$\displaystyle Let u = e^{cost}$

$\displaystyle du=(-e^{cost}sint)dt$

But I don't know what to do from here since it is a sin(2t) in the original integral.

sin(2t) = 2sin(t)cos(t)

do a substitution, u = cos(t), finish off with integration by parts

3. Originally Posted by Em Yeu Anh
$\displaystyle \int_0^{\pi}e^{cost}sin(2t)dt$

Work I have:
$\displaystyle Let u = e^{cost}$

$\displaystyle du=(-e^{cost}sint)dt$

But I don't know what to do from here since it is a sin(2t) in the original integral.

$\displaystyle \sin(2t) =2\sin(t)\cos(t)$ then

$\displaystyle 2\int \sin(t)\cos(t)e^{\cos(t)}dt$

Set $\displaystyle u=\cos(t) \implies du =-\sin(t)dt$ to get

$\displaystyle -2 \int ue^{u}du$ Now use integration by parts

edit too slow

4. Originally Posted by Em Yeu Anh
$\displaystyle \int_0^{\pi}e^{cost}sin(2t)dt$

Work I have:
$\displaystyle Let u = e^{cost}$

$\displaystyle du=(-e^{cost}sint)dt$

But I don't know what to do from here since it is a sin(2t) in the original integral.

$\displaystyle \sin{2t} = 2\sin{t}\cos{t}$.

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

$\displaystyle = -2\int{-\sin{t}\,e^{\cos{t}}\cdot \cos{t}\,dt}$.

Now use integration by parts...

$\displaystyle u = \cos{t}$ so $\displaystyle du = -\sin{t}$

$\displaystyle dv = -\sin{t}\,e^{\cos{t}}$ so $\displaystyle v = e^{\cos{t}}$.

So $\displaystyle -2\int{-\sin{t}\,e^{\cos{t}}\cdot \cos{t}\,dt} = -2\left[\cos{t}\,e^{\cos{t}} - \int{-\sin{t}\,e^{\cos{t}}\,dt}\right]$

$\displaystyle = -2\left[\cos{t}\,e^{\cos{t}} - e^{\cos{t}}\right] + C$

$\displaystyle = 2e^{\cos{t}} - 2\cos{t}\,e^{\cos{t}} + C$.

5. Originally Posted by Em Yeu Anh
$\displaystyle \int_0^{\pi}e^{cost}sin(2t)dt$

Work I have:
$\displaystyle Let u = e^{cost}$

$\displaystyle du=(-e^{cost}sint)dt$

But I don't know what to do from here since it is a sin(2t) in the original integral.

$\displaystyle sin(2t) = 2sin(t)cos(t)$

and $\displaystyle sin(t) = u, cos(t)dt = du$

$\displaystyle \int_0^{\pi}-e^{u}udu$

take the negative sign out of the integrand, and solve the integral using integration by parts:

$\displaystyle u = u, du = du, dv = e^{u}, v = e^{u}$

can you do it now?

6. Originally Posted by Em Yeu Anh
$\displaystyle \int_0^{\pi}e^{cost}sin(2t)dt$

Work I have:
$\displaystyle Let u = e^{cost}$

$\displaystyle du=(-e^{cost}sint)dt$

But I don't know what to do from here since it is a sin(2t) in the original integral.

$\displaystyle \int\limits_0^\pi e^{\cos t}\sin 2t\,dt=$ $\displaystyle 2\int\limits_0^\pi e^{\cos t}\sin t\cos t\,dt=$ (integration by parts) $\displaystyle 2\left(\left[-\cos t\, e^{\cos t}\right]_0^\pi+\int\limits_0^\pi e^{\cos t}\sin t\,dt\right)$ .
I think the answer is $\displaystyle 4e$