Laplace transform and integration together The question, given in the  textbook, is somewhat different. However, I am rephrasing it as follows:  
$$
\frac{2}{\pi}\int_{0}^{\frac{\pi}{2}} \left[ \mathcal L \lbrace \cos(t\cos\theta) \rbrace \right]dt
$$  
The book then uses the standard table of Laplace transform to arrive at the below expression:
$$
\frac{2}{\pi}\int_{0}^{\frac{\pi}{2}} \frac{s}{s^2 + \cos^2\theta} d\theta
$$  
Then, I do not understand what to do. The book has it all messed up.
Can someone please tell me, step-by-step, what to do next ?  
The original question in the book is:  Let
$$f(t) = \frac{1}{\pi} \int_{0}^{\pi}{\cos(t\cos\theta)~ \mathrm d\theta}.$$ 
Prove  that: $$\mathcal L[f(t)] = \frac{1}{\sqrt{s^2+1}}.$$
 A: You want to evaluate 
$$\frac{2}{\pi} s \int_0^{\pi/2} \frac{d\theta}{s^2+\cos^2{\theta}} $$
This may be done through conversion to a contour integral in the complex plane from which we may apply the residue theorem.  First note that we may extend the integration interval to $[0,2 \pi]$ by symmetry.  Let $z=e^{i \theta}$, then $d\theta = -i dz/z$ and $\cos{\theta} = (z+1/z)/2$.  Therefore we have the integral being equal to
$$\frac{s}{i 2 \pi} \oint_{|z|=1} \frac{dz}{z} \frac{1}{s^2+\frac14 (z+z^{-1})^2} = \frac{2 s}{i \pi} \oint_{|z|=1} dz \frac{z}{z^4+(4 s^2+2) z^2+1}$$
The integrand has 4 poles, at $z=\pm i \left ( \sqrt{s^2+1}+s\right)$ and $z=\pm i/\left(\sqrt{s^2+1}+s\right)$.  Note that only the latter two poles are in the interior of the unit circle $|z|=1$, so we need only compute residues at those poles.  Noting that, for a function of the form $p(z)/q(z)$ with simple pole at $z=z_0$, the residue at that pole is $p(z_0)/q'(z_0)$, we find that the integral is, by the residue theorem, the sum of the residues at the poles inside the unit circle, or
$$i 2 \pi \frac{2 s}{i \pi} \frac{2}{-4 (\sqrt{s^2+1}-s)^2+4 (2 s^2+1)} = \frac{1}{\sqrt{s^2+1}}$$
A: The first thing we can note is that $f(t)$ is the Bessel function $J_0(x)$. It is necessary to know that the Bessel function of order $0$ is defined as the solution to: $$x^2y''+xy'+x^2y=0$$
I decided to go about solving the original problem slightly differently.
The Laplace transform of the equation $x(y''+y)+y'=0$ gives us:
\begin{align}0&=-\frac{d}{ds}\mathcal L\{y''+y\}+\mathcal L\{y'\}\\&=-\frac{d}{ds}(s^2\mathcal L\{y\}(s)+\mathcal L\{y\}(s)-sy(0)-y'(0))+s\mathcal L\{y\}(s)-y(0)\\&=-(1+s^2)\mathcal L'\{y\}(s)-s\mathcal L\{y\}(s)\end{align}
We now have $$\mathcal L\{J_0\}(s)=\frac{c}{1+s^2}$$
We can compute $c$ through $$0=\lim_{x\rightarrow\infty}\mathcal L\{J^\prime_0\}(x)=c-1\implies c=1$$
Therefore $$\mathcal L\{f(t)\}=\frac{1}{\sqrt {1+s^2}}$$
