Question

Use integration by parts twice to find the integral of e^(theta)*cos(theta) d(theta)?

Answer 1

let u = \(e^\theta\) and \(dv=\cos\theta ~d\theta\)

Use this to find \(uv - \int v~du \).

You will get

$$
e^\theta \sin\theta + \int e^\theta \sin\theta~d\theta
$$

Do this again, this time with let u = \(e^\theta\) and \(dv=\sin\theta ~d\theta\).

Then you will find that one of the terms will be \(\int e^\theta \cos\theta d\theta\), which is what you started with.

Swing this term back over to the other side of the equation, divide by two. That's your answer.

Answer 2

Your final answer will be:

$$
{e^\theta (\cos\theta+\sin\theta)\over 2}
$$

Answer 3

Make sense?