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| author | Opheliar99 <> | 2010-07-04 02:21:19 +0000 |
|---|---|---|
| committer | bnewbold <bnewbold@adelie.robocracy.org> | 2010-07-04 02:21:19 +0000 |
| commit | 32e82967016f48f091427bc4f57706aba5194d01 (patch) | |
| tree | 489365632e0d48a87bd18a179319c0a76f48c780 /Problem Set 2.page | |
| parent | 1d9ba4013d87ddeb4a1404f5c9812b2ebf13c417 (diff) | |
| download | afterklein-wiki-32e82967016f48f091427bc4f57706aba5194d01.tar.gz afterklein-wiki-32e82967016f48f091427bc4f57706aba5194d01.zip | |
posted solutions of 2 and 3 in pset2
Diffstat (limited to 'Problem Set 2.page')
| -rw-r--r-- | Problem Set 2.page | 2 |
1 files changed, 1 insertions, 1 deletions
diff --git a/Problem Set 2.page b/Problem Set 2.page index 5d48dd5..1672b23 100644 --- a/Problem Set 2.page +++ b/Problem Set 2.page @@ -40,7 +40,7 @@ $\int_0^{2\pi} |\sin^2(x)|^2 dx = \sum |a_n|^2.$ 2. Since $\sin x = \frac{e^{ix}-e^{-ix}}{2}$, -$\int_0^{2\pi} \sin^4(x) dx = \frac{{\left e^{ix}-e^{-ix} \right}^{4}}{16}$, +$\int_0^{2\pi} \sin^4(x) dx = \frac{(\left e^{ix}-e^{-ix})}^{4}}{16}$, $ = \frac{e^{i 4x}+e^{-i 4x}-4 e^{i 2x} -4 e^{-i 2x}+6}{16}$. |