1. ## definite integral!

$\displaystyle \int_{-\pi}^{\pi} \cos^{2}ny dy$

$\displaystyle \left[\frac{n\cos^{3}ny}{3n}\sin{ny}\right]_{-\pi}^{\pi}$

$\displaystyle \left[\frac{n\cos^{3}n\pi}{3n}\sin{n\pi}+\frac{n\cos^{3} n\pi}{3n}\sin{n\pi}\right]$

$\displaystyle 2\left[\frac{n\cos^{3}n\pi}{3n}\sin{n\pi}\right]$

$\displaystyle 2\left[0\right]$

pls is 0 equivalent to $\displaystyle 2{\pi}$ or $\displaystyle \pi$

thanks!

2. ## Re: definite integral!

Originally Posted by lawochekel
$\displaystyle \int_{-\pi}^{\pi} \cos^{2}ny dy$

$\displaystyle \left[\frac{n\cos^{3}ny}{3n}\sin{ny}\right]_{-\pi}^{\pi}$
You have do this incorrectly. See here.

3. ## Re: definite integral!

Hello, lawochekel!

You integrated incorrectly.

We need the identity: .$\displaystyle \cos^2\theta \:=\:\tfrac{1}{2}\left(1 + \cos2\theta\right)$

$\displaystyle \int_{\text{-}\pi}^{\pi} \cos^2(ny)\,dy$

$\displaystyle \int_{\text{-}\pi}^{\pi} \cos^2(ny)\,dy \;=\;\tfrac{1}{2}\int^{\pi}_{\text{-}\pi}\big[1 + \cos(2ny)\big]\,dy$

. . . . . . . . . . . $\displaystyle =\; \tfrac{1}{2}\big[y + \tfrac{1}{2n}\sin(2ny)\bigg]^{\pi}_{\text{-}\pi}$

. . . . . . . . . . . $\displaystyle =\;\tfrac{1}{2}\left[\pi + \tfrac{1}{2n}\sin(2\pi n)\right] - \tfrac{1}{2}\left[\text{-}\pi + \tfrac{1}{2n}\sin(\text{-}2\pi n)\right]$

. . . . . . . . . . . $\displaystyle =\;\tfrac{1}{2}(\pi + 0) - \tfrac{1}{2}(\text{-}\pi + 0)$

. . . . . . . . . . . $\displaystyle =\; \pi$