Evaluating (Laplace) integral

Question

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?

Answer 1

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 $x=\tan \theta$

Answer 2

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.