Trouble integrating square of a sine 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:
$$\begin{align} &\frac{2}{L} \int^{\frac{L}{3}}_{\frac{L}{6}} \sin^2 kx dx \\
 &= \frac{1}{L} \int^{\frac{L}{3}}_{\frac{L}{6}} (1 - \cos 2kx) dx \\
 &= \frac{1}{L} \left[ x - \frac{1}{2k} \sin2kx \right]^{\frac{L}{3}}_{\frac{L}{6}} \\
 &= \frac{1}{3} + \frac{1}{L} \left[- \frac{L}{2 \pi} \sin \frac{2 \pi x}{L} \right]^{\frac{L}{3}}_{\frac{L}{6}} \\
 &= \frac{1}{3} - \left[\frac{1}{2 \pi} \sin \frac{2 \pi}{x} \right]^{3}_{6} \\
 &= \frac{1}{3} + \frac{\sqrt{3}}{2 \pi} \neq \frac{1}{6} \end{align}$$
 A: 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 $1/6$.
A: 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
$$
\int^{\frac{L}{3}}_{\frac{L}{6}} \sin^2(kx) ~\mathrm{d}x = \frac{1}{2} \left(\frac{L}{3}-\frac{L}{6}\right).
$$
