# Inserting Limits Into Integral Result

• Jan 10th 2009, 12:43 PM
Simplicity
Inserting Limits Into Integral Result
What would this give: $\left[\frac{\cos n\pi t}{n \pi}\right]^0_{-1}$.

The solution given is $(1 + \cos n\pi)$.

I don't understand what to do with the $n\pi$ because $\cos(0) = 1$ but what does $\cos n\pi (0)$ equal? (Worried)
• Jan 10th 2009, 12:50 PM
galactus

$cos(n{\pi})=-1$ for odd values of n

$cos(n{\pi})=1$ for even values of n.

Is that what you meant?.
• Jan 10th 2009, 12:56 PM
Simplicity
Quote:

Originally Posted by galactus

$cos(n{\pi})=-1$ for odd values of n

$cos(n{\pi})=1$ for even values of n.

Is that what you meant?.

$\int^0_{-1} (1+t)\cos n \pi t \ \mathrm{d}t$ Through which I used integration by parts and got a similar result to $\left[\frac{\cos n\pi t}{n \pi}\right]^0_{-1}$.
The correct is $\left[-\frac{\cos n\pi t}{n \pi}\right]^0_{-1}$ but what would I get if I put the limits in?
$\int_{-1}^{0} (1+t)cos(n{\pi}t)dt$
$\frac{1-cos(n\pi)}{(n{\pi})^{2}}$