# Contour Integration & Integration by Parts

I need to find the value of $\displaystyle \int _0^{2\pi}\sin^2 \left(\frac{-\pi}{6}+3e^{it} \right)dt$.

I figured I could use contour integration and the Cauchy-Goursat theorem to do so.

I parametrized the contour using $\gamma(t)=\frac{-\pi}{6}+3e^{it}$ and had the function be $f(z)=\sin^2z$.

$f(z)$, then is an entire function and is thus analytic on the trace and in the interior of the simple closed contour parametrized by $\gamma(t)$.

Then, I could use contour integration to

$\displaystyle \int_C \sin^2(z)dz=\displaystyle \int_0^{2\pi}\sin^2 \left(\frac{-\pi}{6}+3e^it \right)\cdot3ie^{it}dt=0$ due to Cauchy-Goursat.

How can I use the above equation to solve for $\displaystyle \int _0^{2\pi}\sin^2 \left(\frac{-\pi}{6}+3e^{it} \right)dt$? I'm stuck and not sure where to go from here.

You can get appropriately sized parentheses (or in fact any paired delimiters) by preceding them with \left and \right. – joriki Apr 26 '13 at 6:02
With $z=\mathrm e^{\mathrm it}$ and $\mathrm dz=\mathrm i\mathrm e^{\mathrm it}\mathrm dt$ and thus $\mathrm dt=-\mathrm i\mathrm e^{-\mathrm it}\mathrm dz=-\mathrm i\mathrm z^{-1}\mathrm dz$ we have
$$\int_0^{2\pi}\sin^2\left(\frac{-\pi}6+3\mathrm e^{\mathrm it}\right)\mathrm dt=-\mathrm i\oint\sin^2\left(\frac{-\pi}6+3z\right)z^{-1}\mathrm dz\;.$$
The residue at $z=0$ is $\sin^2(-\pi/6)=1/4$, so the integral is $-\mathrm i\cdot2\pi\mathrm i\cdot1/4=\pi/2$.