definite integral!

**lawochekel** · September 21st 2012, 02:22 AM

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

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

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

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

$2\left[0\right]$

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

thanks!

**Plato** · September 21st 2012, 02:40 AM

Originally Posted by lawochekel

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

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

You have do this incorrectly. See here.

**Soroban** · September 21st 2012, 07:32 AM

Hello, lawochekel!

You integrated incorrectly.

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

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

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

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

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

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

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

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