How to evaluate $\int_{-\pi/2}^{\pi/2} \tan x \cos (A \cos x +B \sin x) \, dx$?

Question

$$\int_{-\pi/2}^{\pi/2} \tan x \cos (A \cos x +B \sin x) \, dx$$

Is it possible to calculate this? Both A and B are non-zero and assumed to be real numbers.

I tried Integrate[Tan[x]*Cos[A*Cos[x]+B*Sin[x]],{x,-Pi/2,Pi/2},PrincipalValue->True], but it didn't work.

I would be very grateful if you could share some of the good integration skills, ideas, or any advice.

p.s. I think the integration result should be expressed as a combination of Bessel functions.

You wrote "I think the integration result should be expressed as a combination of Bessel functions". Why do you think so? — user64494, Jun 17 '20 at 05:47
@user64494 I guess it might result from the integral representation of Bessel functions. — Αλέξανδρος Ζεγγ, Jun 17 '20 at 06:42
@Αλέξανδρος Ζεγγ : I don't see any $\tan$ there. Also the bounds of the integration differ. Can you elaborate your comment? — user64494, Jun 17 '20 at 08:01
@user64494 Yes, you are right. There is no tangent there, but a similar part. — Αλέξανδρος Ζεγγ, Jun 17 '20 at 08:02
@Αλέξανδρος Ζεγγ : You words do not correspond to reality. There is no similar part in the integrands there and here. — user64494, Jun 17 '20 at 08:04
@user64494 The real reason must come from the one who made the statement (yes, the OP). I just gave a wild guess according to what I knew. Sorry for any inconvenience caused. ¯_(ツ)_/¯ — Αλέξανδρος Ζεγγ, Jun 17 '20 at 08:09
@user64494 Oh, there is a similar part, in the form of a trigonometric function applied to another trigonometric function in the integrated. BTW, a sum $ a \sin x + b \cos x $ can be merged into a single trigonometric form as $ A \sin(x + \phi) $, you must know I guess. — Αλέξανδρος Ζεγγ, Jun 17 '20 at 08:17
@Αλέξανδρος Ζεγγ : Your words do not correspond to reality again: One sees $\cos(n \tau -x \sin \tau)$ in the definition of $J_n(x)$. — user64494, Jun 17 '20 at 08:22
@Αλέξανδρος Ζεγγ : If you read the question, you will see "Both A and B are non-zero". I definitely end this empty talk. — user64494, Jun 17 '20 at 08:31
@user64494 Yes, but it does not matter. As I have put in one of my comments above, $ A \cos x + B \sin x = C \sin(x + \phi) $, where $ C = \sqrt{A^2 + B^2}, \phi = \arctan(A/B) $. — Αλέξανδρος Ζεγγ, Jun 17 '20 at 08:35

score 5 · Accepted Answer · answered Jun 17 '20 at 21:08

One can find a representation in terms of infinite series

Integrate[Tan[x]Series[Cos[b Sin[x]+a Cos[x]],{a,0,5}]//Normal,{x,-Pi/2,Pi/2},PrincipalValue->True]

It yields $$\frac1{\pi}I= -a J_1(b)+\frac{a^3}{6b} J_2(b)-\frac{a^5}{40b^2}J_3(b)+\ldots$$

We can continue to find the following representation

$$ \frac1{\pi}I=\sum_{i=1}^{\infty} \frac{(-1)^{i}a^{2i-1}}{2^{i-1} (i-1)! (2i-1)b^{i-1}} J_{i}(b). $$

So, you were right about the Bessel functions.

Note

I am cautiously optimistic that a closed form may be found due to the existence of recursive relations that can be used to reduce $J_i$ to a sum of just few Bessel functions of lower order. However, Mathematica seem not to recognize these relations. Maybe someone here can help.

Bravo! I was hacking away at this myself and I didn't think to take the series in $a$. — flinty, Jun 17 '20 at 21:17
@flinty Just wondering if a closed-form expression exists. Replacing all $J_i$ in the sum with a constant leads to the erf-function. But this does not bring me further. — yarchik, Jun 17 '20 at 21:22

How to evaluate $\int_{-\pi/2}^{\pi/2} \tan x \cos (A \cos x +B \sin x) \, dx$?

1 Answers1

Note