# integration question

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• Jan 19th 2009, 07:35 PM
needhelp101
integration question
right now i'm stuck on this integration problem. it's integrating (1/theta^2)cos(1/theta). any suggestions??
• Jan 19th 2009, 07:55 PM
Soroban
Hello, needhelp101!

Quote:

$\displaystyle \int \frac{\cos\left(\frac{1}{\theta}\right)}{\theta^2} \,d\theta$

We have: .$\displaystyle \int \cos\left(\theta^{-1}\right)\,\left(\theta^{-2}d\theta\right)$

Let $\displaystyle u \:=\:\theta^{-1} \quad\Rightarrow\quad du \:=\:-\theta^{-2}d\theta \quad\Rightarrow\quad \theta^{-2}d\theta \:=\:-du$

Substitute: .$\displaystyle \int \cos u\,(-du) \;=\;-\!\!\int\cos u\,du$

Got it?

• Jan 19th 2009, 08:00 PM
needhelp101
yup. i got it now. thanks for the help. i really appreciate it.
• Jan 19th 2009, 08:06 PM
hatsoff
Quote:

Originally Posted by needhelp101
right now i'm stuck on this integration problem. it's integrating (1/theta^2)cos(1/theta). any suggestions??

$\displaystyle \int\theta^{-2}\cos{\theta^{-1}}d\theta$

Substitution:

$\displaystyle u=\theta^{-1}$

$\displaystyle -du=\theta^{-2}d\theta$

$\displaystyle \int\theta^{-2}\cos{\theta^{-1}}d\theta=-\int\cos{u}\;du$

$\displaystyle =-\sin{u}$

$\displaystyle =-\sin{\theta^{-1}}$

Check the solution by taking the derivative:

$\displaystyle -\frac{d}{d\theta}\sin{\theta^{-1}}=\theta^{-2}\cos{\theta^{-1}}$

Looks good!