What is the solution to the following trigonometric integral equation?

Question

What is the solution for $s$ in the following integral equation $$h = \int_{0}^{s}\frac{\mathrm{d}x}{(\sin(x)+c)^3}$$ where $h$ and $c$ are constants.

Maple and Mathematica can compute the integral but then solving for $s$ requires finding a root of a complicated trigonometric expression.

If there are no closed form expressions for $s$ then are there fast and accurate numerical approximations for finding $s$? (I.e. faster than: (1) doing Newton iterations on the resulting non-linear system; and (2) solving the integral using quadrature and stopping when the integral equals $h$)

Thank you!

Answer 1

Since you already have the antiderivative, the equation you need to solve for $s$ is $$h=A+B\tan ^{-1}\left(\frac{c \tan \left(\frac{s}{2}\right)+1}{\sqrt{c^2-1}}\right)+\frac{\cos (s) \left(4 c^2+3 c \sin (s)-1\right)}{2 \left(c^2-1\right)^2 (c+\sin (s))^2}$$ Where $$A=-\frac{\sqrt{c^2-1} \left(4 c^2-1\right)+2 c^2\left(2c^2+1\right) \tan ^{-1}\left(\frac{1}{\sqrt{c^2-1}}\right)}{2 c^2 \left(c^2-1\right)^{5/2}}\qquad B=\frac{\left(2 c^2+1\right) }{\left(c^2-1\right)^{5/2}}$$ This assumes a bunch of conditions $$(s\notin \mathbb{R}\lor \Re(s)\leq \pi )\land ((\Re(c \csc (s))\geq 0\land c \csc (s)\neq 0)\lor$$ $$ c \csc (s)\notin \mathbb{R}\lor \Re(c \csc (s))\leq -1)\land c+\sin (s)\neq 0$$

The rhs presents discontinuities at $(2k+1)\pi$ and we can just focus on the first one.

Now, the problem is with $h$. If it is positive, a look at the curves for any $c > 1$ shows that the range of $h$ is quite limited since the maximum value of the rhs is $$\text{max}=\frac{\sqrt{c^2-1} \left(c \left(2 c \left(\pi c \sqrt{\frac{c^2}{c^2-1}}-4\right)+\pi \sqrt{\frac{c^2}{c^2-1}}\right)+2\right)-2c^2\left(2 c^2+1\right) \cot ^{-1}\left(\sqrt{c^2-1}\right)}{2 c^2 \left(c^2-1\right)^{5/2}}$$ $$\text{min}=\frac{\sqrt{c^2-1} \left(2-c \left(2 c \left(\pi \sqrt{\frac{c^2}{c^2-1}} c+4\right)+\pi \sqrt{\frac{c^2}{c^2-1}}\right)\right)-2c^2 \left(2 c^2+1\right) \cot ^{-1}\left(\sqrt{c^2-1}\right)}{2 c^2 \left(c^2-1\right)^{5/2}}$$ and, in order to have solution, we need ${min} < h < {max}$.

Where the problem is nice is

• if $h > 0$, compute the straight line joining points $(0,0)$ and $(\pi,max)$
• if $h < 0$, compute the straight line joining points $(\pi,min)$ and $(2\pi,0)$

Interpolating along these lines will give a starting guess for Newton method.

Trying for $c=2$ for which $max=\frac{4 \sqrt{3} \pi -15}{36}$ and $min=-\frac{15+8 \sqrt{3} \pi}{36}$

Being lazy, I started at the midepoint

• for the first case, I used $h=\frac 34 max$ and the iterates are $$\left( \begin{array}{cc} n & s_n \\ 0 & 1.57080 \\ 1 & 2.83934 \\ 2 & 2.64193 \\ 3 & 2.61501 \\ 4 & 2.61462 \end{array} \right)$$
• for the second case, I used $h=\frac 34 min$ and the iterates are

$$\left( \begin{array}{cc} n & s_n \\ 0 & 4.71239 \\ 1 & 4.44687 \\ 2 & 4.43695 \\ 3 & 4.43691 \end{array} \right)$$

This seems to work quite well.

Update

Assuming $h > 0 $, we can do a few more things. First, name $f(s)$ the rhs of the equation in the top.

Better than what I wrote earlier, the maximum value is given by $$f(\pi)=\frac{1-4 c^2}{c^2 \left(c^2-1\right)^2}+\frac{\left(2 c^2+1\right) \sec ^{-1}(c)}{\left(c^2-1\right)^{5/2}}$$

The interesting properties are listed in the table below

$$\left( \begin{array}{cccc} s & f(s) & f'(s) & f''(s) \\ 0 & 0 & \frac{1}{c^3} & -\frac{3}{c^4} \\ \frac{\pi }{2} & \frac{1}{2}f(\pi ) & \frac{1}{(c+1)^3} & 0 \\ \pi & f(\pi ) & \frac{1}{c^3} & \frac{3}{c^4} \end{array} \right)$$ which (for sure) shows the symmetry at $s=\frac \pi 2$. This could probably be the base information for some interpolating function (as a polynomial, it would be of quite high degree and then no interest). Another possible solution would be to use the data in the above table to fit a cubic equation in $s$ without the constant term (this is how looks $f(s)$ in $[0,\pi]$) but the problem is to decide about the respective weights to be assigned to the function and derivative values. I am still working on it but it would then require solving the cubic in $s$ using the hyperbolic method.

For the time being, using linear interpolation between the end points $$s_0=\frac {\pi}{f(\pi)} h$$ For the worked case $\big[c=2 , h=\frac 34 f(\pi)\big]$, this will give $s_0=\frac {3 \pi} 4$ and Newton iterates would then be $$\left( \begin{array}{cc} n & s_n \\ 0 & 2.35619 \\ 1 & 2.64598 \\ 2 & 2.61514 \\ 3 & 2.61462 \end{array} \right)$$

Ignoring the values of the second derivative and giving the same weights to the other values, the cubic polynomial $\alpha s+\beta s^2+\gamma s^3$was generted minimizing the sum of the squred residuals. The formulae are really messy and will not be printed here.

For the worked example, the result from the cubic would be $s=2.62062$.

New update

For a better estimate, for $0 \leq h \leq \frac{1}{2}f(\pi )$, using the following data

$$\left( \begin{array}{cccc} s & f(s) & f'(s) & f''(s) \\ 0 & 0 & \frac{1}{c^3} & \\ \frac{\pi }{2} & \frac{1}{2}f(\pi ) & \frac{1}{(c+1)^3} & 0 \end{array} \right)$$

we can write the rhs as $$f(s)\approx\frac{\alpha s+\beta s^2}{ 1+\gamma s+ \delta s^2}$$ and the parameters $(\alpha,\beta,\gamma,\delta)$ are totally explicit (the formulae are easy to establish) and the estimate is obtained at the price of a quadratic equation in $s$.

Furthermore, it is worth to mention that, using this with the simplistic $s_0$, a single iteration leads almost to the solution. Let $s_{(n)}$ the result for the worked example.

$$\left( \begin{array}{ccc} n & s_{(n)} & \text{method} \\ 1 & 2.6459758 & \text{Newton} \\ 2 & 2.6130746 & \text{Halley} \\ 3 & 2.6140472 & \text{Householder} \\ 4 & 2.6148101 & \text{no name} \\ 5 & 2.6146006 & \text{no name} \end{array} \right)$$

Answer 2

I prefer to write a separate answer.

Using the tangent half-angle substitution $x=2 \tan ^{-1}(t)$, we have $$I=\int\frac{dx}{(\sin(x)+c)^3}=2\int \frac{ \left(t^2+1\right)^2}{\left(c t^2+2 t+c\right)^3}\,dt$$ Partial fraction decomposition leads to $$f(t)=\int \frac{ \left(t^2+1\right)^2}{\left(c t^2+2 t+c\right)^3}\,dt=\frac{\left(2-c^2\right) t+c}{c^3 \left(c^2-1\right) \left(c t^2+2 t+c\right)^2}+$$ $$\frac{c\left(5 c^2-2\right) t+(4 c^4-3 c^2+2) } {2 c^3 \left(c^2-1\right)^2 \left(c t^2+2 t+c\right) }+\frac{\left(2 c^2+1\right) }{2 \left(c^2-1\right)^{5/2}}\tan ^{-1}\left(\frac{c t+1}{\sqrt{c^2-1}}\right)$$ making that we need to solve $$\frac h 2+f(0)=f\left(\tan \left(\frac{s}{2}\right)\right)$$ which is not bad.