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@@ -39,8 +39,8 @@ $\int_0^{2\pi} |\sin^2(x)|^2 dx = \sum |a_n|^2.$
# Solutions
2. Since
-$\sin(x) = \frac{\exp^{ix}-e^{-ix}}{2}$,
-$\int_0^{2\pi} \sin^4(x) dx = \frac{{e^{ix}-e^{-ix}}^4}{16}$,
+$\sin x = \frac{e^{ix}-e^{-ix}}{2}$,
+$\int_0^{2\pi} \sin^4(x) dx = \frac{{e^{ix}-e^{-ix}}^{4}}{16}$,
$ = \frac{e^{i 4x}+e^{-i 4x}-4 e^{i 2x} -4 e^{-i 2x}+6}{16}$
If we express any periodic function $f(x)$ as