Solve: $\int_a^b\frac{1}{x\sqrt{x^2-a^2}}\left(\frac{\cos(x)}{x}+\sin(x)\right)dx$ I am trying to solve this integral $$\int_a^b\frac{1}{x\sqrt{x^2-a^2}}\left(\frac{\cos(x)}{x}+\sin(x)\right)dx.$$
I have tried some basic substitution and integration by parts, but beyond that can seem to solve find an analytical solution. I was able to solve it numerically for which an example solution is shown in the picture with the integral on the $y$ axis and $x$ on the $x$ axis. It looks like a Bessel or sinusoidal exponential to me.

 A: Adding to my comment I found an answer. I will explain the technique but leave it to you on how to evaluate it. So the integral we are required to evaluate is,
$$\int_{a}^{b}\frac{\cos(x)}{x^{2}\sqrt{x^{2}-a^{2}}}\, \mathrm dx$$
Using the series expansion of cosine
$$\sum_{k\ge 0}\frac{(-1)^{k}}{(2k)!}\int_{a}^{b}\frac{x^{2k-2}}{\sqrt{x^{2}-a^{2}}}\,\mathrm dx$$
Put $x=au \implies \mathrm dx=a\,\mathrm du$
$$\sum_{k\ge 0}\frac{(-1)^{k}a^{2k-2}}{(2k)!}\int_{1}^{\frac{b}{a}}\frac{u^{2k-2}}{\sqrt{u^{2}-1}}\,\mathrm du $$
Now do a simple substitution,
Let $u=\cosh(\theta) \implies \mathrm du=\sinh(\theta)\,\mathrm d\theta$.
Then the integral will simplify to,
$$\sum_{k\ge 0}\frac{a^{2k-2}(-1)^{k}}{(2k)!}\int_{0}^{A}\cosh^{2k-2}(\theta)\,\mathrm d\theta$$
Where $A$ denotes the $\operatorname{arcosh}(\frac{b}{a})$.
Write $\cosh(\theta)=\frac{e^{\theta}+e^{-\theta}}{2}$ and use the binomial theorem ( break the sum at $0$ and continue from 1, because at 0 binomial theorem will not work.) So after we will be left with only summation term and integral will be solved. Summation will look like,
$$\sum_{k\ge 1}\frac{a^{2k-2}(-1)^{k}}{(2k)!}\sum_{u=0}^{2k-2}\binom{2k-2}{u}2^{2-2k}\left(\frac{e^{A(2u-2k+2)}-1}{2u-2k+2}\right)$$
(I have ignored the value at $k=0$, because the important and dominating term will be mentioned summation)
