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| -rw-r--r-- | Problem Set 2.page | 5 |
1 files changed, 3 insertions, 2 deletions
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 |