On the wikipedia's article about integration using Euler's formula
they use $\displaystyle\int\mathrm{\sin}^{2}x\cos{4x}\,\mathrm{d}x$ as an example:
$$\displaystyle\int\mathrm{\sin}^{2}x\cos{4x}\,\mathrm{d}x$$
$$\displaystyle\int\mathrm{\bigg(\frac{e^{ix}-e^{-ix}}{2i}\bigg)^2\bigg(\frac{e^{4ix}+e^{-4ix}}{2}\bigg)}\,\mathrm{d}x$$
$$\displaystyle-\frac{1}{8}\int\mathrm{e^{6ix}+e^{-6ix}-2e^{4ix}-2e^{-4ix}+e^{2ix}+e^{-2ix}}\,\mathrm{d}x$$
And then it says that we can either integrate as it is or substitute the integrand with $\cos 6x - 2 \cos 4x + \cos2x$, shouldn't that be $2\cos 6x - 4 \cos 4x + 2\cos2x$? Is wikipedia wrong or what am I not understanding?
Answer
Rewrite the integral as
$$
\begin{align}
&-\frac{1}{8}\int\left(\mathrm{e^{6ix}+e^{-6ix}-2e^{4ix}-2e^{-4ix}+e^{2ix}+e^{-2ix}}\right)\,\mathrm{d}x
\\&=-\int\left(\mathrm{\frac14\cdot\frac{e^{6ix}+e^{-6ix}}{2}-\frac24\cdot\frac{e^{4ix}+e^{-4ix}}{2}+\frac14\cdot\frac{e^{2ix}+e^{-2ix}}{2}}\right)\,\mathrm{d}x\\
&=-\int\left(\frac14\cos6x-\frac12\cos4x+\frac14\cos2x\right)\,\mathrm{d}x\\
&=-\frac14\int\left(\color{blue}{\cos6x-2\cos4x+\cos2x}\right)\,\mathrm{d}x\\
\end{align}
$$
but
$$
\begin{align}&-\int\left(\frac14\cos6x-\frac12\cos4x+\frac14\cos2x\right)\,\mathrm{d}x
\\&=-\frac18\int\left(\color{red}{2\cos6x-4\cos4x+2\cos2x}\right)\,\mathrm{d}x.
\end{align}$$
I think you are right.
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