Inserting Limits Into Integral Result

**Simplicity** · January 10th 2009, 12:43 PM

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?

**galactus** · January 10th 2009, 12:50 PM

What is your question?.

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

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

Is that what you meant?.

**Simplicity** · January 10th 2009, 12:56 PM

Originally Posted by galactus

What is your question?.

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

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

Is that what you meant?.

I had integral of:

$\int^0_{-1} (1+t)\cos n \pi t \ \mathrm{d}t$ 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?

**galactus** · January 10th 2009, 01:01 PM

$\int_{-1}^{0} (1+t)cos(n{\pi}t)dt$

The result should be:

$\frac{1-cos(n\pi)}{(n{\pi})^{2}}$

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