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I have the following expression which I am trying to evaluate:

$$ \frac{2}{L} \int^{\frac{L}{3}}_{\frac{L}{6}} \sin^2 kx$$ where $k = \frac{\pi}{L}$.

According to my calculator, the answer should be $\frac{1}{6}$ but I can't seem to get this result by manual integration.

Here's what I've tried so far:

$$\begin{align} &\frac{2}{L} \int^{\frac{L}{3}}_{\frac{L}{6}} \sin^2 kx dx \\ &= \frac{1}{L} \int^{\frac{L}{3}}_{\frac{L}{6}} (1 - \cos 2kx) dx \\ &= \frac{1}{L} \left[ x - \frac{1}{2k} \sin2kx \right]^{\frac{L}{3}}_{\frac{L}{6}} \\ &= \frac{1}{3} + \frac{1}{L} \left[- \frac{L}{2 \pi} \sin \frac{2 \pi x}{L} \right]^{\frac{L}{3}}_{\frac{L}{6}} \\ &= \frac{1}{3} - \left[\frac{1}{2 \pi} \sin \frac{2 \pi}{x} \right]^{3}_{6} \\ &= \frac{1}{3} + \frac{\sqrt{3}}{2 \pi} \neq \frac{1}{6} \end{align}$$

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2 Answers 2

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There are two errors: First, $$ \frac{1}{L} \bigl[ x \bigr]^{\frac{L}{3}}_{\frac{L}{6}} = \frac 16 $$ and not $1/3$. Second, $$ \left[\frac{1}{2 \pi} \sin \frac{2 \pi}{x} \right]^{3}_{6} = \frac{1}{2 \pi} \left( \sin \frac{2\pi}{3}- \sin \frac{ \pi}{3}\right) = 0 \, . $$ It seems that you added the terms instead of subtracting them.

With these corrections you'll get the expected result $1/6$.

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There is a trick. Integrate by parts $$ \int^{\frac{L}{3}}_{\frac{L}{6}} \sin^2(kx)~\mathrm{d}x = -\frac{1}{k}\cos(kx)\sin(kx) \bigg \vert^{\frac{L}{3}}_{\frac{L}{6}} + \int^{\frac{L}{3}}_{\frac{L}{6}} \cos^2(kx)~\mathrm{d}x = \int^{\frac{L}{3}}_{\frac{L}{6}} \cos^2(kx)~\mathrm{d}x $$ So: $$ \left(\frac{L}{3}-\frac{L}{6}\right) = \int^{\frac{L}{3}}_{\frac{L}{6}} 1~\mathrm{d}x = \int^{\frac{L}{3}}_{\frac{L}{6}} \sin^2(kx) + \cos^2(kx)~\mathrm{d}x = $$ $$ \int^{\frac{L}{3}}_{\frac{L}{6}} \sin^2(kx) ~\mathrm{d}x+ \int^{\frac{L}{3}}_{\frac{L}{6}} \cos^2(kx)~\mathrm{d}x =2\int^{\frac{L}{3}}_{\frac{L}{6}} \sin^2(kx) ~\mathrm{d}x $$ Rearrange to get $$ \int^{\frac{L}{3}}_{\frac{L}{6}} \sin^2(kx) ~\mathrm{d}x = \frac{1}{2} \left(\frac{L}{3}-\frac{L}{6}\right). $$

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