Approximate Trigonometric Integral

Question

Let the function $I(x)$ be defined as $$I(x) = \int_{-\pi/2}^{\pi/2}{\rm d} \theta \ \frac{\cos^2{\theta}}{\sqrt{\cos^2\theta+x^2\sin^2\theta}} \ , $$

Then we have for instance $I(0)=2$. Is there a nice answer for $I(x)$ for $x$ small? The most obvious thing to do would be to binomially expand $$I(x) \approx \int_{-\pi/2}^{\pi/2}{\rm d} \theta \ \left( \cos \theta - \frac{x^2}{2}\frac{\sin^2\theta}{\cos\theta}\right) \ , $$

but this doesn't seem promising since the $\cos \theta$ in the denominator of the second term renders it no longer small near $\pm \pi/2$. For what it's worth, Mathematica gives an answer in terms of elliptic functions for $I(x)$ so I'm expecting $I(x)$ to be smooth enough that some approximation should exist.

Answer 1

Substitute $t = \cot \theta$. Then

$$ I(x) = 2 \int_{0}^{\infty} \frac{t^2}{\sqrt{t^2+x^2} (1+t^2)^{3/2}} \, \mathrm{d}t. $$

Now using the antiderivative

$$ 2 \int \Biggl( \frac{t^2}{\sqrt{t^2+x^2}} - t \Biggr) \, \mathrm{d}t = x^2\biggl( \frac{t}{t+\sqrt{t^2+x^2}} - \log\bigl(t + \sqrt{t^2 + x^2}\bigr) \biggr) + C, $$

we get

$$ I(x) = 2 + x^2 \log x + 3x^2 \int_{0}^{\infty} \biggl( \frac{t}{t+\sqrt{t^2+x^2}} - \log\bigl(t + \sqrt{t^2 + x^2}\bigr) \biggr) \frac{t}{(1+t^2)^{5/2}} \, \mathrm{d}t. $$

This proves that

$$ I(x) = 2 + x^2 \log x + \mathcal{O}(x^2) $$

as $x \to 0^+$. Studying the above formula would reveal more detailed expansion for $I(x)$.

Answer 2

There is something interesting : starting from the result given by Wofram Alpha $$I(x)=\frac{2 \left(x^2 K\left(1-x^2\right)-E\left(1-x^2\right)\right)}{x^2-1}$$ Expanding as a series around $x=0$ $$I(x)=2 +\sum_{n=1}^\infty \Big[a_n-b_n \log(2)+c_n \log(x^2)\Big]\, x^{2n}$$ and the coefficients are $$\left( \begin{array}{cccc} n & a_n & b_n & c_n \\ 1 & \frac{3}{2} & 2 & \frac{1}{2} \\ 2 & \frac{51}{32} & \frac{9}{4} & \frac{9}{16} \\ 3 & \frac{105}{64} & \frac{75}{32} & \frac{75}{128} \\ 4 & \frac{40985}{24576} & \frac{1225}{512} & \frac{1225}{2048} \\ 5 & \frac{110439}{65536} & \frac{19845}{8192} & \frac{19845}{32768} \\ 6 & \frac{2224761}{1310720} & \frac{160083}{65536} & \frac{160083}{262144} \end{array} \right)$$ where interesting patterns seem to appear.

Using the terms in the table, there is a good fit for $0 \leq x \leq \frac 34$ (the maximum error being $0.014$ at the right bound; it is only $6.92 \times 10^{-5}$ for $x=\frac 12$.