# Homework Help: Integration by Parts

1. Dec 4, 2009

1. The problem statement, all variables and given/known data

Evaluate the integral
(e^-theta) cos(2theta)

I got this as my answer
e^(-theta)-sin(2theta)+cos(2theta)e^(-theta)+C
But it was wrong
All help is appreciated.

2. Dec 4, 2009

### rock.freak667

well you don't have "1/2" in there anywhere

u=e so du=?

dv= cos2θ dθ so v= ?

3. Dec 4, 2009

i got du= e^-theta
and v= sin(2theta)

4. Dec 4, 2009

### rock.freak667

v would be "1/2 sin2θ", right?

5. Dec 4, 2009

This is probably a stupid question but why would it be?

6. Dec 4, 2009

### rock.freak667

$$\int cos n\theta d\theta$$

let t = nθ ⇒ dt = ndθ or dt/n = dθ (n is a constant)

$$\therefore \int cos n\theta d\theta \equiv \int \frac{cos t}{n} dt = \frac{1}{n} sin t=\frac{1}{n}sin(n\theta)$$

see where the 1/2 comes from?

7. Dec 4, 2009

ok yeah i think i got it, so now I'm at
e^-θ - sin(2θ) - ∫1/2sin(2θ) * e^-θ
what do i do with the ∫1/2sin(2θ) * e^-θ?
Take the anti derivative right?
would that be -(1/2)cos(2θ) * e^-θ?

8. Dec 4, 2009

### rock.freak667

integrate by parts again

9. Dec 5, 2009

### HallsofIvy

What is the derivative of $sin(2\theta)$?

In general, if $\int f(x)dx= F(x)+ C$ then to integrate $\int f(ax+b) dx$, let u= ax+ b so that du= a dx or (1/a)du= dx. The integral becomes $(1/a)\int f(u)du= (1/a)F(u)+ C= (1/a)F(ax+ b)+ C$.

That is, for f(ax+b), just as, if you were differentiating, you would have to multiply by a (by the chain rule), so when integrating, you divide by a.

(That works for a simple linear substitution.