Given $ \int_c{\frac{sin^2z}{(z-\frac{\pi}{6})^3}dz}$

An $\lvert z \lvert = 1$ , is a circle with centre at the origin and radius (r) is unity. The pole $ Z_0 = \frac{\pi}{6} $ lies inside the circle 'c'.

$\therefore f(z) = sin^2z $ is analytic in and on 'c', hence by cauchy theorem

$\begin{align} \int_c{\frac{f(z)}{(z-z_0)^n}} = \frac{2 \pi i}{(n-1)!}f^{n-1}(z_0) \\ \int_c{\frac{sin^2z}{(z-\frac{\pi}{6})^3}dz}= \frac{2 \pi i}{(3-1)!}f^{3-1}(z_0) \\ \int_c{\frac{sin^2z}{(z-\frac{\pi}{6})^3}dz}= \frac{2 \pi i}{2}f^{2}(z_0) \\ \int_c{\frac{sin^2z}{(z-\frac{\pi}{6})^3}dz}= \pi if''(z_0) ------1 \\ \end{align}$

Where $f(z) = sin^2z $

Now, $f'(z) = 2sinz.cosz$

$\begin{align} f"(z) = 2cosz.cosz + 2sinz(-sinz) \\ = 2cos^2z - 2sin^2z \\ \end{align}$

$ at z_0 = \frac{\pi}{6}$

$\begin{align} f"(z_0) = f"(\frac{\pi}{6}) = 2cos^2(\frac{\pi}{6}) - 2sin^2(\frac{\pi}{6}) \\ f"(z_0) = 2(\frac{3}{4}) - 2(\frac{1}{4}) \\ = (\frac{3}{2}) - (\frac{1}{2}) \\ f"(z_0) = 1 - - - - 2 \\ put \ equation \ 2 \ in \ equation \ 1 \\ \int_c{\frac{sin^2z}{(z-\frac{\pi}{6})^3}dz} = \pi i (1) \\ \int_c{\frac{sin^2z}{(z-\frac{\pi}{6})^3}dz} = \pi i \\ \end{align}$