Have I done this correctly?
Evaluate
$\displaystyle\int_{0}^{2\pi}\sin^3(3e^{i\theta} +\frac{\pi}{4})d\theta$
Gauss MVT:
$$f(z_0)=\frac{1}{2\pi}\displaystyle\int_0^{2\pi}f(z_0+re^{i\theta})d\theta$$
So we have the following:
$$\frac{1}{2\pi}f(z_0)=\frac{1}{2\pi}\displaystyle\int_0^{2\pi} \sin^3(3e^{i\theta} + \frac{\pi}{4})d\theta $$
With,
$$z_0=\frac{\pi}{4}$$ and
\begin{align*}
f(z_0) &= sin^3(z_0) \\
f(\frac{\pi}{4})&=\frac{1}{2\sqrt2} \\
\frac{1}{2\pi}f(z_0)&=\frac{1}{2\pi}\cdot \frac{1}{2\sqrt2} \\
&= \frac{1}{4\sqrt{2}\pi}\\
\end{align*}
Thanks.
$$
\begin{align}
\int_0^\pi\log(\sin(x))\,\mathrm{d}x
&=\int_0^\pi\log\left(\color{#C00000}{\frac12}\sqrt{2-2\cos(x)}\right)\,\mathrm{d}x\\
&=\pi\color{#00A000}{\frac1{2\pi}\int_0^{2\pi}\log\left(\sqrt{2-2\cos(x)}\right)\,\mathrm{d}x}\color{#C00000}{-\pi\log(2)}\\[6pt]
&=\pi\color{#00A000}{\log(1)}-\pi\log(2)\\[12pt]
&=-\pi\log(2)
\end{align}
$$
Best Answer
Since $\sin^3 z$ is a holomorphic function, its average over a circle centered at $\pi/4$ is equal to $\sin^3(\pi/4)=2^{-3/2}$. The average is $\frac{1}{2\pi}\int_0^{2\pi}\dots$, therefore $$ \int_{0}^{2\pi}\sin^3(3e^{i\theta} +\frac{\pi}{4})\,d\theta = 2\pi \cdot 2^{-3/2}$$