I am developing an algorithm and am stuck with the following definite integral:
$$\int_s^t\frac{\sin\theta \cos\theta}{((a^2-1)\cos^2\theta+1)^2}\sin\left(\frac{\pi(\theta-b+c)}{2c}\right)\,d\theta$$
Without the second sine term it would be very simple, however in this form I am getting nowhere. The parameters satisfy $s,t,b,c\in[0,\frac{\pi}{2}]$, $a\ge0$.
I'd be most happy with a solution or even better, a simple analytic approximation that is easy to compute in time critical applications.
Any help is greatly appreciated.
Thank you
if $c$ is small then the right $\sin$ will oscillate quickly causing problems for quadrature methods.
The left trig functions have slow oscillation so over a short range $s,t \in [0,\frac{\pi}{2}]$ it will have half a cycle. Note: $\cos(\theta)\sin(\theta) = \frac1{2}\sin(2\theta)$, $\displaystyle 2\frac{\pi}{2} = \pi$, i.e. half a cycle and it saves one trig calculation.
The left trig functions can be approximated accurately by polynomials over the short range of integration. This is required for many quadrature methods.
There are many papers on the integration of oscillating functions.
$\sin(x) = Imaginary(e^{ix})$
The problem can be expressed in the form: $$I = \int_a^b f(x)e^{iwx}dx$$
where $f(x)$ is the slow left trig terms and $e^{iwx}$ is the fast right $\sin$.
This form can be transformed into a Gauss–Laguerre quadrature form. $\int_{0}^{\infty} f(u)e^{-u}du$
An integral path is taken into the complex plane.
$$h_x(t) = x + it$$
$x$ is fixed and $t$ varies from $0\dots\infty$.
$$I = \int_0^\infty f(h_x(t))e^{iwh_x(t)}h'_x(t) dt \, \Biggr|_{x=b}^{x=a} $$
$$I = ie^{iwx}\int_0^\infty f(x+it)e^{-wt} dt \, \Biggr|_{x=b}^{x=a} $$
Let $u = wt$ then $t = \frac{u}{w}$ and $dt = \frac{du}{w}$
$$I = \frac{ie^{iwx}}{w}\int_0^\infty f\left(x+i\frac{u}{w}\right)e^{-u} du \, \Biggr|_{x=b}^{x=a} $$
The integral is now in Gauss Laguerre form.
Determine the number of quadrature points for the required accuracy.
Calculate the quadrature using the points and weights then take the imaginary component.