Proof for $I_0(1)=\int_{0}^{\pi}e^{\cos x}\ \mathrm{d}x<\frac{3\pi}{2}$. I want to prove
$$\int_{0}^{\pi}e^{\cos x}\ \mathrm{d}x<\frac{3\pi}{2}.$$
Here I can provide a proof as follows:
\begin{align*}
\int_{0}^{\pi}e^{\cos x}\ \mathrm{d}x
 &=\int_{0}^{\pi}\sum_{n=0}^{\infty}\frac{\cos^nx}{n!}\ \mathrm{d}x\\
 &=\sum_{n=0}^{\infty}\frac{1}{n!}\int_{0}^{\pi}\cos^nx\ \mathrm{d}x\\
 &=\sum_{n=0}^{\infty}\frac{1}{(2n)!}\int_{0}^{\pi}\cos^{2n}x\ \mathrm{d}x\\
 &=\sum_{n=0}^{\infty}\frac{2}{(2n)!}\int_{0}^{\frac\pi2}\cos^{2n}x\ \mathrm{d}x\\
 &=\pi+\sum_{n=1}^{\infty}\frac{2}{(2n)!}\cdot\frac{(2n-1)!!}{(2n)!!}\cdot\frac{\pi}{2}\\
 &=\pi\left(1+\sum_{n=1}^{\infty}\frac{1}{((2n)!!)^2}\right)\\
 &<\pi\left(1+\sum_{n=1}^{\infty}\frac{1}{2^{n+1}}\right)\\
 &=\frac{3\pi}{2}.
\end{align*}
Is there a more elmentary and easy proof. I know that
$$I_0(x)=\int_{0}^{\pi}e^{x\cos t}\ \mathrm{d}t$$
is the $0$th modified Bessel function of the first kind.
 A: Denote your integral by $I$. Taking $x = \pi  - t$, we see that
$$
I = \int_0^\pi  {{\rm e}^{ - \cos t} {\rm d}t} .
$$
Thus by averaging,
$$
I = \int_0^\pi  {\cosh (\cos x){\rm d}x}  = 2\int_0^{\pi /2} {\cosh (\cos x){\rm d}x} .
$$
Then by Taylor's formula with Lagrange error bound
$$
I \le 2\int_0^{\pi /2} {{\rm d}x}  + 2\frac{{\cosh (1)}}{2}\int_0^{\pi /2} {\cos ^2 x\,{\rm d}x}  = \pi  + \cosh (1)\frac{\pi }{4}.
$$
Here
$$
\cosh (1) = \frac{{{\rm e} + 1/{\rm e}}}{2} \le \frac{{3 + 1}}{2} = 2.
$$
A: Estimating the convex exponential function from above by secants on $[-1, 0]$ and $[0, 1]$ gives
$$
 e^u \le \begin{cases}
 1 + (e-1)u & \text{ for } 0 \le u \le 1 \, ,\\
 1 + (1-1/e)u & \text{ for } -1 \le u \le 0 \, .
\end{cases}
$$
It follows that
$$
 \int_0^\pi e^{\cos x} \, dx \le \int_0^{\pi/2} (1+(e-1)\cos x)\, dx
+ \int_{\pi/2}^\pi (1+(1-1/e)\cos x) \, dx \\
= \pi + e + \frac 1e - 2 \approx 4.22775
$$
and that is less than $3\pi/2 \approx 4.712389$.
A: First split the integral as follows :
$$\int_{0}^{\pi}e^{\cos x}\ \mathrm{d}x =\int_{0}^{\frac{\pi}{2}}e^{\cos x}\ \mathrm{d}x +  \int_{\frac{\pi}{2}}^{\pi}e^{\cos x}\ \mathrm{d}x \\ =  \int_{0}^{\frac{\pi}{2}}e^{\cos x}\ \mathrm{d}x +  \int_{0}^{\frac{\pi}{2}}e^{-\sin x}\ \mathrm{d}x \\ = \int_{0}^{\frac{\pi}{2}}e^{\cos x}\ \mathrm{d}x +  \int_{0}^{\frac{\pi}{2}}e^{-\cos x}\ \mathrm{d}x \\ = 2 \int_{0}^{\frac{\pi}{2}}\cosh(\cos x) \ \mathrm{d}x \\ \leq 2\epsilon \cosh(1) + (\pi - 2 \epsilon) \cosh(\cos\epsilon) \ \forall \epsilon \in \left[0,\frac{\pi}{2}\right] $$
Taking $\epsilon = \pi/4$, we get that :
$$ \leq \frac{\pi}{2} \left(\cosh 1 + \cosh\left(\frac{1}{\sqrt 2}\right)\right) \lt 3\pi/2$$
