$\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.
Thanks in advance!
$\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.
Thanks in advance!
$\displaystyle \sin{2t} = 2\sin{t}\cos{t}$.
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 $\displaystyle -2\int{-\sin{t}\,e^{\cos{t}}\cdot \cos{t}\,dt} = -2\left[\cos{t}\,e^{\cos{t}} - \int{-\sin{t}\,e^{\cos{t}}\,dt}\right]$
$\displaystyle = -2\left[\cos{t}\,e^{\cos{t}} - e^{\cos{t}}\right] + C$
$\displaystyle = 2e^{\cos{t}} - 2\cos{t}\,e^{\cos{t}} + C$.
$\displaystyle sin(2t) = 2sin(t)cos(t)$
and $\displaystyle sin(t) = u, cos(t)dt = du$
so your equation now is:
$\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?
$\displaystyle \int\limits_0^\pi e^{\cos t}\sin 2t\,dt=$ $\displaystyle 2\int\limits_0^\pi e^{\cos t}\sin t\cos t\,dt=$ (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 $\displaystyle 4e$
Tonio