right now i'm stuck on this integration problem. it's integrating (1/theta^2)cos(1/theta). any suggestions??
Hello, needhelp101!
$\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?
$\displaystyle \int\theta^{-2}\cos{\theta^{-1}}d\theta$
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!