From 1d9ba4013d87ddeb4a1404f5c9812b2ebf13c417 Mon Sep 17 00:00:00 2001
From: Opheliar99 <>
Date: Sun, 4 Jul 2010 02:20:32 +0000
Subject: posted solutions of 2 and 3 in pset2

---
 Problem Set 2.page | 5 +++--
 1 file changed, 3 insertions(+), 2 deletions(-)

(limited to 'Problem Set 2.page')

diff --git a/Problem Set 2.page b/Problem Set 2.page
index d001972..5d48dd5 100644
--- a/Problem Set 2.page	
+++ b/Problem Set 2.page	
@@ -40,8 +40,9 @@ $\int_0^{2\pi} |\sin^2(x)|^2 dx = \sum |a_n|^2.$
 
 2. Since
 $\sin x = \frac{e^{ix}-e^{-ix}}{2}$,
-$\int_0^{2\pi} \sin^4(x) dx = \frac{{e^{ix}-e^{-ix}}^{4}}{16}$,
-$ = \frac{e^{i 4x}+e^{-i 4x}-4 e^{i 2x} -4 e^{-i 2x}+6}{16}$
+$\int_0^{2\pi} \sin^4(x) dx = \frac{{\left e^{ix}-e^{-ix} \right}^{4}}{16}$,
+
+$ = \frac{e^{i 4x}+e^{-i 4x}-4 e^{i 2x} -4 e^{-i 2x}+6}{16}$.
 
 If we express any periodic function $f(x)$ as 
 
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