1. ## Integration by Parts

Hi, can anyone help me solve this:

First make a substitution then use integration by parts to evaluate the integral:

$\displaystyle \int^\pi_0 e^{\cos{t}}\sin{2t} dt$

So far I have:

$\displaystyle \int^\pi_0 e^{\cos{t}}2\sin{t}\cos{t} dt$

let u = $\displaystyle \cos{t}$

How do I do the next part?

2. Originally Posted by coldfire
Hi, can anyone help me solve this:

First make a substitution then use integration by parts to evaluate the integral:

$\displaystyle \int^\pi_0 e^{\cos{t}}\sin{2t} dt$

So far I have:

$\displaystyle \int^\pi_0 e^{\cos{t}}2\sin{t}\cos{t} \, dt$

let u = $\displaystyle \cos{t}$

How do I do the next part?
$\displaystyle du = -\sin{t} \, dt$

$\displaystyle -2 \int^\pi_0 e^{\cos{t}}(-\sin{t})\cos{t} dt$

$\displaystyle -2 \int_1^{-1} e^u \cdot u \, du$

$\displaystyle 2 \int_{-1}^{1} e^u \cdot u \, du$

now do parts

3. Originally Posted by coldfire
Hi, can anyone help me solve this:

First make a substitution then use integration by parts to evaluate the integral:

$\displaystyle \int^\pi_0 e^{\cos{t}}\sin{2t} dt$

So far I have:

$\displaystyle \int^\pi_0 e^{\cos{t}}2\sin{t}\cos{t} dt$

let u = $\displaystyle \cos{t}$

How do I do the next part?
Spoiler:

Then $\displaystyle du = -\sin t \ dt$. Your integral went from $\displaystyle t=0$ to $\displaystyle t=\pi$; now it should go from $\displaystyle u = \cos 0 = 1$ to $\displaystyle u = \cos \pi = -1$.

so

$\displaystyle I = -2\int_1^{-1} e^{u}u\ du$

Now we use integration by parts with $\displaystyle dv = e^u$, $\displaystyle w=u$. Then

$\displaystyle I = -2\left((ue^u)^{u=-1}_{u=1}-\int_1^{-1} e^{u}\ du\right)$

$\displaystyle =-2\left((-e^{-1}-e)-(e^{-1}-e)\right)$

$\displaystyle =4e^{-1}$

4. Thanks guys.