# How do you integrate inte^(cos)(t) (sin 2t) dt between a = 0, b = pi?

May 14, 2015

This integral has to be done with the theorem of the integration by parts, that says:

$\int f \left(x\right) g ' \left(x\right) \mathrm{dx} = f \left(x\right) g \left(x\right) - \int g \left(x\right) f ' \left(x\right) \mathrm{dx}$.

We have to remember also the double-angle formula of sinus, that says:

$\sin 2 \alpha = 2 \sin \alpha \cos \alpha$.

So, the integral becomes:

$2 {\int}_{0}^{\pi} {e}^{\cos} t \sin t \cos t \mathrm{dt} = - 2 {\int}_{0}^{\pi} {e}^{\cos} t \left(- \sin t\right) \cdot \cos t \mathrm{dt} = \left(1\right)$

We can assume that:

$f \left(t\right) = \cos t$ and $g ' \left(t\right) \mathrm{dx} = {e}^{\cos} t \left(- \sin t\right) \mathrm{dt}$.

So:

$f ' \left(t\right) \mathrm{dt} = - \sin t \mathrm{dt}$

$g \left(t\right) = {e}^{\cos} t$ and this is because $\int {e}^{f} \left(x\right) f ' \left(x\right) \mathrm{dx} = {e}^{f} \left(x\right) + c$.

Then:

$\left(1\right) = - 2 \left\{{\left[\cos t \left({e}^{\cos} t\right)\right]}_{0}^{\pi} - {\int}_{0}^{\pi} {e}^{\cos} t \left(- \sin t\right) \mathrm{dt}\right\} =$

$= - 2 {\left[\cos t {e}^{\cos} t - {e}^{\cos} t\right]}_{0}^{\pi} =$

$= - 2 \left(\cos \pi {e}^{\cos} \pi - {e}^{\cos} \pi - \left(\cos 0 {e}^{\cos} 0 - {e}^{\cos} 0\right)\right) =$

$= - 2 \left(- {e}^{-} 1 - {e}^{-} 1 - e + e\right) = \frac{4}{e}$.