Evaluating (Laplace) integral I wish to evaluate the following integral:
$$f(t)=\frac{1}{\pi}\int_{0}^{\pi}\cos\left(t\sin\theta\right)\,d\theta$$
I started with the 'standard' form of the Laplace Transform:
$$\mathcal{L}(\cos \omega t) = \frac{s}{s^2 + \omega ^2}$$
Now I can substitute to obtain:
$$\mathcal{L}[f(t)] = \frac{1}{\pi}\int_{0}^{\pi} \frac {s}{s^2 + \sin^2 \theta} d\theta $$
But how do I get back? Taking the 's' out:
$$\mathcal{L}[f(t)] = \frac{s}{\pi}\int_{0}^{\pi} \frac {d \theta}{s^2 + \sin^2 \theta}$$
But how to proceed now?
 A: If you divide by $\cos^2 \theta$ in numerator and denominator, then,
$$\mathcal{L}[f(t)] = \frac{s}{\pi}\int_{0}^{\pi} \frac {\sec^2 \theta d \theta}{\sec^2 \theta *s^2 + \tan^2 \theta}$$
$$\implies \mathcal{L}[f(t)] = \frac{s}{\pi}\int_{0}^{\pi} \frac {\sec^2 \theta}{s^2 + (1+s^2)\tan^2 \theta} d \theta$$
Since, $d(\tan \theta)=\sec^2 \theta$, the further part is easy to solve.
Edited for query in comment:
$$ \mathcal{L}[f(t)] = \frac{s}{\pi}\int_{0}^{\pi} \frac {\sec^2 \theta}{s^2 + (1+s^2)\tan^2 \theta} d \theta $$
which can be transformed to the form $$ \int \frac {1}{a^2 + x^2} dx=\frac {1}{a}\tan^{-1}\frac{x}{a}+c $$
by substituting $x=\tan \theta$
A: Those integrals can be seen in various places around the web. From this link we find that
$$J_n(x)=\frac1{\pi}\int_0^{\pi}\cos(n\tau-x\sin\tau)d\tau$$
And also we have the Laplace transform
$$\mathcal{L}\left\{J_n(\omega t)\right\}=\frac{\left(\sqrt{s^2+\omega^2}-s\right)^n}{\omega^n\sqrt{s^2+\omega^2}}$$
And I've seen
$$\int_0^{\pi}\frac{d\theta}{a^2\sin^2\theta+b^2\cos^2\theta}=\frac{\pi}{ab}$$
Come up a lot in this forum.
