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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.

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    $\begingroup$ What are the conditions on $f$ in order that the mean value theorem be satisfied? Does your function satisfy them? $\endgroup$ – Ron Gordon Mar 5 '13 at 17:55
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    $\begingroup$ I think that the answer is not true. The correct answer must be f(z0)*2pi. $\endgroup$ – user79057 May 23 '13 at 2:37
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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 $$ \int_{0}^{2\pi}\sin^3(3e^{i\theta} +\frac{\pi}{4})\,d\theta = 2\pi \cdot 2^{-3/2}$$

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