# integration question

• Jan 19th 2009, 08:35 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, 08:55 PM
Soroban
Hello, needhelp101!

Quote:

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

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

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

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

Got it?

• Jan 19th 2009, 09:00 PM
needhelp101
yup. i got it now. thanks for the help. i really appreciate it.
• Jan 19th 2009, 09:06 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??

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

Substitution:

$u=\theta^{-1}$

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

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

$=-\sin{u}$

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

Check the solution by taking the derivative:

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

Looks good!