Integral $\int_{-\infty}^{\infty} \frac{\cos t \sin( \sqrt{1+t^2})}{\sqrt{1+t^2}}dt$ I want to evaluate the integral $$\int_{-\infty}^{\infty} \frac{\cos t \sin( \sqrt{1+t^2})}{\sqrt{1+t^2}}dt$$
I tried using Feynman’s trick and introduced the parameter $e^{-\sqrt{1+t^2} x}$, then differentiating with respect to $x$ under the integral. However I don’t see a good way to evaluate the resulting integral. Is there a contour method? 
This was an AMM problem ($12145$). 

Another solution for an interesting antiderivative.
$$I=\int\dfrac{\cos\left(t\right)\sin\left(\sqrt{t^2+1}\right)}{\sqrt{t^2+1}}\,dt$$
First write
$$I=\frac12\int\frac{\sin \left(\sqrt{t^2+1}+t\right)}{\sqrt{t^2+1}}\,dt+\frac12\int\frac{\sin \left(\sqrt{t^2+1}-t\right)}{\sqrt{t^2+1}}\,dt$$
For the first integral, use $u=\sqrt{t^2+1}+t$ to get by the end
$$\int\frac{\sin \left(\sqrt{t^2+1}+t\right)}{\sqrt{t^2+1}}\,dt=\int \frac {\sin(u)} u \,du=\text{Si}(u)=\text{Si}(\sqrt{t^2+1}+t)$$ For the second integral, use $u=\sqrt{t^2+1}-t$ to get by the end
$$\int\frac{\sin \left(\sqrt{t^2+1}-t\right)}{\sqrt{t^2+1}}\,dt=-\int \frac {\sin(u)} u \,du=-\text{Si}(u)=-\text{Si}(\sqrt{t^2+1}-t)$$
So
$$I=\frac 12\Big(\text{Si}\left(\sqrt{t^2+1}+t\right)-\text{Si}\left(\sqrt{t^2+1}-t\right)\Big)$$ For the definite integral
$$J=\int_{-p}^p\dfrac{\cos\left(t\right)\sin\left(\sqrt{t^2+1}\right)}{\sqrt{t^2+1}}\,dt=\text{Si}\left(p-\sqrt{p^2+1}\right)+\text{Si}\left(p+\sqrt{p^2+1}\right)$$
Now, for large values of $p$, composing Taylor series
$$p-\sqrt{p^2+1}=-\frac{1}{2 p}+\frac{1}{8 p^3}+O\left(\frac{1}{p^5}\right)$$
$$\text{Si}\left(p-\sqrt{p^2+1}\right)=-\frac{1}{2 p}+\frac{19}{144 p^3}+O\left(\frac{1}{p^5}\right)\qquad \to 0$$
$$p+\sqrt{p^2+1}=2 p+\frac{1}{2 p}-\frac{1}{8 p^3}+O\left(\frac{1}{p^5}\right)$$
Now, using for large values of $a$
$$\text{Si}(a)=\frac{\pi}{2} -\frac{\left(a^2-2\right) \cos (a)}{a^3}-\frac{\sin (a)}{a^2}+\cdots$$ then the result.
Using $p=100$, the above truncated expressions lead to $1.5733$ while the numerical integration gives $1.5634$.
