For $\alpha \in \mathbb{R}$, define $\displaystyle I(\alpha):=\int_{0}^{2\pi}e^{\alpha \cos \theta}\cos(\alpha \sin \theta)\; d\theta$. Calculate $I(0)$. Hence evaluate $\displaystyle\int_{0}^{2\pi}e^{\cos \theta}\cos( \sin \theta)\; d\theta$.
Hint: To evaluate the integral that expresses $\displaystyle\frac{dI}{d\alpha}$, consider $\displaystyle\frac{\partial}{\partial \theta}(e^{\alpha \cos \theta}\sin(\alpha \sin \theta))$.
How do I do this question? I think this might have something to do with the Fundamental Theorem of Calculus, but I'm not sure.
I computed $\displaystyle I(0)=\int_{0}^{2\pi} d\theta=2 \pi$, and $\displaystyle I(1)=\int_{0}^{2\pi}e^{\cos \theta}\cos( \sin \theta) d\theta$. Following the hint I get
$$\begin{align}
\frac{\partial}{\partial \theta}(e^{\alpha \cos \theta}\sin(\alpha \sin \theta)) & =\alpha e^{\alpha \cos \theta} \sin (\alpha \sin \theta) + e^{\alpha \cos \theta}\cos(\alpha \sin \theta) \alpha \cos \theta \\
& = \alpha e^{\alpha \cos \theta} \sin (\alpha \sin \theta) + \frac{dI}{d \alpha} \cos \theta. \\
\end{align}$$
Is this correct so far?
The answers in the question referred as a duplicate does not help. I'm in a course dealing with real values, not complex.
$$
\begin{align}
\int_0^\pi\log(\sin(x))\,\mathrm{d}x
&=\int_0^\pi\log\left(\color{#C00000}{\frac12}\sqrt{2-2\cos(x)}\right)\,\mathrm{d}x\\
&=\pi\color{#00A000}{\frac1{2\pi}\int_0^{2\pi}\log\left(\sqrt{2-2\cos(x)}\right)\,\mathrm{d}x}\color{#C00000}{-\pi\log(2)}\\[6pt]
&=\pi\color{#00A000}{\log(1)}-\pi\log(2)\\[12pt]
&=-\pi\log(2)
\end{align}
$$
Best Answer
First a correction:
$$\begin{align} \frac{\partial}{\partial \theta}(e^{\alpha \cos \theta}\sin(\alpha \sin \theta)) & =-\alpha \sin \theta \, e^{\alpha \cos \theta} \sin (\alpha \sin \theta) + e^{\alpha \cos \theta}\cos(\alpha \sin \theta) \alpha \cos \theta \\ \end{align}$$
Now \begin{align} \frac{dI}{d\alpha}&=\frac{d}{d\alpha}\int_{0}^{2\pi}e^{\alpha \cos \theta}\cos(\alpha \sin \theta) d\theta \\ &=\int_{0}^{2\pi}\frac{d}{d\alpha}(e^{\alpha \cos \theta}\cos(\alpha \sin \theta)) d\theta \\ &=\int_{0}^{2\pi}\cos \theta \, e^{\alpha \cos \theta}\cos(\alpha \sin \theta)- e^{\alpha \cos \theta}\sin(\alpha \sin \theta)\sin \theta \, d\theta \\ &=\int_{0}^{2\pi}\frac{1}{\alpha} \frac{\partial}{\partial \theta}(e^{\alpha \cos \theta}\sin(\alpha \sin \theta)) d\theta \\ &=\frac{1}{\alpha} \Big[e^{\alpha \cos \theta}\sin(\alpha \sin \theta)\Big]_0^{2\pi} \\ &=0 \end{align}
So $I(\alpha)$ is actually constant.
So $I(1)=I(0)=2\pi$
So the answer is $2\pi$