How to evaluate $\int_{0}^{2\pi}e^{\cos \theta}\cos( \sin \theta) d\theta$? 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.
 A: I simplified my approach.

As a generalization of Venus' answer , assume that $\alpha$ is a positive real number and that $f(z)$ is an entire function.
Then $$ \begin{align} \int_{0}^{2 \pi} \left(  f(\alpha e^{i\theta})+f(\alpha e^{-i \theta}) \right) \, \mathrm  d \theta &= \int_{0}^{2 \pi} \left(\sum_{n=0}^{\infty} \frac{f^{(n)}(0)}{n!} (\alpha e^{i \theta})^{n} + \sum_{n=0}^{\infty}\frac{f^{(n)}(0)}{n!} (\alpha e^{-i \theta})^{n} \right) \, \mathrm d \theta \\ &= 2\int_{0}^{2 \pi} \sum_{n=0}^{\infty} \frac{f^{(n)}(0)}{n!} \alpha^{n} \cos(n \theta) \, \mathrm d \theta \\ &= 2\sum_{n=0}^{\infty} \frac{f^{(n)}(0)}{n!} \, \alpha^{n} \int_{0}^{2 \pi} \cos (n \theta) \, \mathrm d \theta \\ &= 2 f(0) \int_{0}^{2 \pi} \, d \theta + 2 \sum_{n=1}^{\infty}\frac{f^{(n)}(0)}{n!} \, \alpha^{n}  \int_{0}^{2 \pi} \cos (n \theta) \, \mathrm d \theta \\ &=4 \pi f(0)  + 2\sum_{n=1}^{\infty}\frac{f^{(n)}(0)}{n!} (0) \\ &= 4 \pi f(0). \end{align}$$
If we let $f(z) = e^{z}$,  we get $$ 2 \int_{0}^{2 \pi} e^{\alpha \cos \theta} \cos (\alpha \sin \theta) \, \mathrm d \theta = 4 \pi (1) = 4 \pi.$$
And if we let $f(z) = \cos(z)$, for example, we get $$2 \int_{0}^{2 \pi} \cos (\alpha \cos \theta) \cosh(\alpha \sin \theta)  \, \mathrm d \theta = 4 \pi (1) = 4 \pi.$$
A: 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 $2\pi$
A: Alternatively, we know that $$\Re\left( e^{\Large e^{i\theta}}\right)=e^{\cos\theta}\cos(\sin\theta)$$
Using Taylor series of exponential function and Euler's formula we have $$e^{\Large e^{i\theta}}=1+(\cos\theta+i\sin\theta)+\frac{(\cos2\theta+i\sin2\theta)}{2!}+\frac{(\cos3\theta+i\sin3\theta)}{3!}+\cdots$$
Using $\displaystyle\int_0^{2\pi}\cos(n\theta)\;d\theta=0$ for $n$ is integer and $n\neq0$, we get
$$\begin{align}\int_0^{2\pi}e^{\cos\theta}\cos(\sin\theta)\;d\theta&=\int_0^{2\pi}\Re\left( e^{\Large e^{i\theta}}\right)d\theta\\&=\int_0^{2\pi}\left(1 +\cos\theta+\frac{\cos2\theta}{2!}+\frac{\cos3\theta}{3!}+\cdots\right)d\theta\\&=2\pi\end{align}$$
A: Here is a solution using real analysis, different from solutions already published (involving derivation under integral sign, series, etc.). This solution is based on a connection with cardinal sine function.
Let
$$f(x):=e^{-\cos x} \cos(\sin x)$$
It is periodic function with the following graphical representation:

As it is symmetrical with respect to vertical line $x=\pi$, it is sufficient to be able to compute
$$I_1:=\int_0^{\pi/2} f(x)dx \ \ \text{and} \ \ I_2:=\int_{\pi/2}^{\pi} f(x)dx$$
The result will be $2(I_1+I_2)$.
Preliminary result:
$$\int^{\color{red}{\pi/2}}_0 e^{-p\cos x} \cos(p\sin x)\,dx=\int^{\infty}_p  \frac{\sin(t)}{t}dt\tag{1}$$
which is formula NT 13(26) page 486 of my 1980 edition of Gradshteyn and Ryzhik.
(where we recognize the opposite of a primitive function of cardinal sine function).


*

*Integral $I_1$ has the following expression by taking $p=-1$ in (1):

$$I_1=\int^{\infty}_{-1}  \frac{\sin(t)}{t}dt\tag{2}$$

*

*It is now sufficient to establish that :

$$I_2=\int_{-\infty}^{-1}  \frac{\sin(t)}{t}dt\tag{3}$$
In this way, we will have $2(I_1+I_2)=2\int_{-\infty}^{+\infty}  \frac{\sin(t)}{t}dt=2 \pi$
by using the classical result $\int_{-\infty}^{\infty} \dfrac{\sin(x)}{x}dx=\pi.$
Here is the proof of (3):
$$I_2:=\int_{\pi/2}^{\pi} f(x)dx \ \ \overset{x:=\pi-X}{=} \ \ -\int_{\pi/2}^{0} e^{-\cos(X)}\cos(\sin(X))dX$$
$$I_2=\int_0^{\pi/2} e^{-\cos(X)}\cos(\sin(X))dX$$
Using (1) with $p=1$:
$$I_2=\int_1^{\infty}  \frac{\sin(t)}{t}dt$$
It remains to take the change of variable $T=-t$ to obtain the desired result (3).
A: Rewrite
$$
\int_0^{\large2\pi} e^{\cos\theta}\cos(\sin\theta)\ d\theta=\Re\left[\int_0^{\large2\pi} e^{\Large e^{i\theta}}d\theta\right].
$$
Let
$$
I(\alpha)=\int_0^{\large2\pi} e^{\Large\alpha e^{i\theta}}d\theta,
$$
then
$$
\frac{dI}{d\alpha}=I'(\alpha)=\int_0^{\large2\pi} e^{i\theta}e^{\Large\alpha e^{i\theta}}d\theta.
$$
Rewrite
$$
I'(\alpha)=\frac{1}{i\alpha}\int_0^{\large2\pi} i\alpha e^{i\theta}e^{\Large\alpha e^{i\theta}}d\theta.
$$
Let $x=\alpha e^{i\theta}\;\color{blue}{\Rightarrow}\;dx=i\alpha e^{i\theta}\ d\theta$, then
$$
I'(\alpha)=\frac{1}{i\alpha}\left[e^{\Large\alpha e^{i\theta}}\right]_{\theta=0}^{\large2\pi}=0.
$$
Thus $\Re\left[I'(\alpha)\right]=0$ and $I(\alpha)$ is a constant.
Taking $\alpha=0$ yields $I(0)=2\pi$. Hence $\color{blue}{I(\alpha)=2\pi}$ and consequently
$$
\int_0^{\large\pi} e^{\cos\theta}\cos(\sin\theta)\ d\theta=\large\color{blue}{2\pi}.
$$
A: Let:
$\displaystyle \tag*{} I(n,t) = \int \limits_{0}^{2 \pi} e^{t \cos \theta} \cos( n \theta - t\sin \theta) \mathrm{d \theta}$
$\displaystyle \tag*{} I(n,1) = \int \limits_{0}^{2 \pi} e^{ \cos \theta} \cos( n \theta - \sin \theta) \mathrm{d \theta}$
On differentiating both the sides, we have:
$\displaystyle \tag*{} I'(n,t) = \int \limits _{0}^{2 \pi} \dfrac{\partial( e^{t \cos \theta} \cos( n \theta - t\sin \theta))}{\partial t} \mathrm{d \theta}$
We have:
$\displaystyle \tag*{} \dfrac{\partial( e^{t \cos \theta} \cos( n \theta - t\sin \theta))}{\partial t} = e^{t \cos \theta} \left [(\cos (n  \theta - t\sin \theta))(\cos \theta)  + (\sin (n \theta - t \sin \theta))(\sin \theta)\right ]$
Using the basic trigonometric identity, which states,
$\displaystyle \tag*{} \cos (A-B) = \cos A \cos B + \sin A \sin B$
We obtain:
$\displaystyle \tag*{}  \dfrac{\partial( e^{t \cos \theta} \cos( n \theta - t\sin \theta))}{\partial t} = e^{t \cos \theta} \cos ((n-1) \theta - t \sin \theta)$
Notice that:
$\displaystyle \tag{1} I'(n,t) = I(n-1,t)$
Now, let’s derive some useful solutions
$\displaystyle \tag{2} I(0,0) = \cos(0)\int \limits _{0}^{2 \pi} \mathrm{d \theta} = 2 \pi$
and
$\displaystyle \tag{3} I(n,0) = \int \limits _{0}^{2 \pi} \cos (n \theta) \mathrm{d \theta} = 0$
Now, note that:
$\displaystyle \tag*{} \int \limits _{0}^{t} I'(n,t) = \int \limits _{0}^{t} I(n-1,a) \mathrm{ da}$
$\displaystyle \tag*{} I(n,t) - \underbrace{I (n,0)}_{=0} = \int \limits _{0}^{t} I(n-1,a) \mathrm{ da}$
$\displaystyle \tag{4} I(n,t) = \int \limits _{0}^{t} I(n-1,a) \mathrm{ da}$
Now, to find $I'(0,t)$, to find this, I am going to rename the variable to apply differentiation under integral once again!
Let:
$\displaystyle \tag*{} I(s) = \int \limits _{0}^{2 \pi} e^{s \cos \theta} \cos (-s \sin \theta) \mathrm{ d \theta}$
$\displaystyle \tag*{} I'(s) = \int \limits _{0}^{2 \pi} \dfrac{\partial (e^{s \cos \theta}\cos (-s \sin \theta))}{\partial s} \mathrm{d \theta}$
$\displaystyle \tag*{} \dfrac{\partial (e^{s \cos \theta}\cos (s \sin \theta))}{\partial s} \mathrm{d \theta} = e^{s \cos \theta} \cos (\theta - s \sin \theta)$
and
$\displaystyle \tag*{} e^{s \cos \theta} \cos (\theta - s \sin \theta) \cdot s = \dfrac{\partial(e^{s \cos \theta} \sin (s \sin \theta))}{\partial \theta}$
$\displaystyle \tag*{} I(s) = \dfrac{1}{s} \int \limits _{0} ^{2 \pi} \partial( e^{s \cos \theta} \sin ( s \sin \theta)) = 0$
Now, so $I'(0,t) = 0$ , so why not plug this into $(4)$, we get:
$\displaystyle \tag*{} 
\begin{align} I(0,t)&=2\pi \\\\ I(1,t) &= 2\pi\cdot t \\\\
I(2,t) &= 2 \pi \cdot \dfrac{t^2}{2!} \\\\
 I(n,t) &= 2\pi \cdot \dfrac{t^n}{t!} \end{align}$
and
$\displaystyle \tag*{} I(n,1) = 2 \pi \cdot \dfrac{1^n}{t!}$
Hence,
$\displaystyle \tag*{} \boxed{\boxed{ \int \limits_{0}^{2 \pi} e^{ \cos \theta} \cos( n \theta - \sin \theta) \mathrm{d \theta} = \dfrac{2 \pi}{n!}}}$
Now, since $\cos x$ is even function, we have $\cos(-\sin x) = \cos(\sin x)$. So we have:
$\displaystyle \tag*{} \boxed{\boxed{ \int \limits_{0}^{2 \pi} e^{ \cos \theta} \cos(- \sin \theta) \mathrm{d \theta}=2 \pi= \int \limits_{0}^{2 \pi} e^{ \cos \theta} \cos(\sin \theta) \mathrm{d \theta}}}$
