# Thread: Trig Integration by Parts Problem

1. ## Trig Integration by Parts Problem

$\displaystyle \int3\theta\cos\thetad\theta = [3\theta][\sin\theta] - \int[\sin\theta][3]$

$\displaystyle u = 3\theta$

$\displaystyle dv = \cos\theta d\theta$

$\displaystyle v = \sin\theta$

$\displaystyle du = 3$

$\displaystyle 3\sin^{2}\theta - \int3\sin\theta$

$\displaystyle 3\sin^{2}\theta + 3\cos\theta + C$ - ?? Online homework says this is wrong.

$\displaystyle 3(\sin^{2}\theta + \cos \theta) + C$ - ?? Online homework says this is also wrong.

2. ## Re: Trig Integration by Parts Problem

I agree with your choices of u and dv (though I would put \displaystyle \begin{align*} \mathrm{d}\theta \end{align*} at the end of the \displaystyle \begin{align*} \mathrm{d}u \end{align*} term.

So you need to remember that integration by parts is as follows: \displaystyle \begin{align*} \int{ u\,\mathrm{d}v} = u\,v - \int{ v\,\mathrm{d}u} \end{align*}, so that means with your choices of \displaystyle \begin{align*} u \end{align*} and \displaystyle \begin{align*} \mathrm{d}v \end{align*} you should get \displaystyle \begin{align*} \int{ 3\theta\cos{(\theta)}\,\mathrm{d}\theta} = 3\theta\sin{(\theta)} - \int{ 3\sin{(\theta)}\,\mathrm{d}\theta} \end{align*}. Go from here.

3. ## Re: Trig Integration by Parts Problem

$\displaystyle \int3\theta\cos\thetad\theta = [3\theta][\sin\theta] - \int[\sin\theta][3] d\theta$

$\displaystyle u = 3\theta$

$\displaystyle dv = \cos\theta d\theta$

$\displaystyle v = \sin\theta$

$\displaystyle du = 3 d\theta$

$\displaystyle 3\sin^{2}\theta - \int3\sin\theta$

Next step??

4. ## Re: Trig Integration by Parts Problem

NO! \displaystyle \begin{align*} u\,v = 3\theta\sin{(\theta)} \end{align*} NOT \displaystyle \begin{align*} 3\sin^2{(\theta)} \end{align*}.

Surely you can integrate \displaystyle \begin{align*} 3\sin{(\theta)} \end{align*}...

5. ## Re: Trig Integration by Parts Problem

The answer came out right on homework using ProveIt's suggestion.