Integrate e^(-theta)cos(2theta): Get Help Now!

[addmeup]

Homework Statement

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.

 

[rock.freak667]

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

u=e^-θ so du=?

dv= cos2θ dθ so v= ?

 

[addmeup]

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

 

[rock.freak667]

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

v would be "1/2 sin2θ", right?

 

[addmeup]

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

 

[rock.freak667]

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

[tex]\int cos n\theta d\theta[/tex]

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

[tex]\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)[/tex]


see where the 1/2 comes from?

 

[addmeup]

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?

 

[rock.freak667]

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?

integrate by parts again

 

[HallsofIvy]

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

What is the derivative of [itex]sin(2\theta)[/itex]?

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

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.