Definite integral with hyperbolic cosine and circumference segment

Question

I've been struggling with this definite integral: $$I=\int_{0}^{a}\frac{\sqrt{1-\frac{x^2}{a^2}}}{b+\cosh x} dx,$$ with $$ a>0,\\ b>1.$$

Mathematica won't solve it

Any idea on how to solve this kind of integral?

Answer 1

Even $b=0$ will probably make the problem impossible to solve and I guess that numeical integration could be the only way.

For an approximate result, we could expand the integrand as a Taylor series built at $x=0$ to get $$\frac{\sqrt{1-\frac{x^2}{a^2}}}{b+\cosh (x)}=\frac{1}{b+1}-\frac{x^2 \left(a^2+b+1\right)}{2 \left(a^2 (b+1)^2\right)}-\frac{x^4 \left(a^4 (b-5)-6 a^2 (b+1)+3 (b+1)^2\right)}{24 \left(a^4 (b+1)^3\right)}+\frac{x^6 \left(a^6 (-((b-28) b+61))+15 a^4 (b-5) (b+1)+45 a^2 (b+1)^2-45 (b+1)^3\right)}{720 a^6 (b+1)^4}+O\left(x^8\right)$$ and integrate termwise.

For the definite integral this would lead to $$\int_0^a\frac{\sqrt{1-\frac{x^2}{a^2}}}{b+\cosh (x)}=-\frac{a \left(a^6 \left(b^2-28 b+61\right)+27 a^4 \left(b^2-4 b-5\right)+543 a^2 (b+1)^2-4029 (b+1)^3\right)}{5040 (b+1)^4}+\cdots$$ which could be acceptable for small values of $a$ as shown below $$\left( \begin{array}{cccc} a & b & \text{approximation} & \text{exact} \\ \frac{1}{4} & 1 & 0.099507 & 0.097793 \\ \frac{1}{4} & 2 & 0.066431 & 0.065280 \\ \frac{1}{4} & 3 & 0.049858 & 0.048992 \\ \frac{1}{4} & 4 & 0.039903 & 0.039209 \\ & & & \\ \frac{1}{3} & 1 & 0.132247 & 0.129999 \\ \frac{1}{3} & 2 & 0.088382 & 0.086864 \\ \frac{1}{3} & 3 & 0.066368 & 0.065223 \\ \frac{1}{3} & 4 & 0.053134 & 0.052215 \\ & & & \\ \frac{1}{2} & 1 & 0.196565 & 0.193344 \\ \frac{1}{2} & 2 & 0.131756 & 0.129550 \\ \frac{1}{2} & 3 & 0.099089 & 0.097412 \\ \frac{1}{2} & 4 & 0.079403 & 0.078050 \\ & & & \\ 1 & 1 & 0.375025 & 0.370035 \\ 1 & 2 & 0.255071 & 0.251336 \\ 1 & 3 & 0.193296 & 0.190346 \\ 1 & 4 & 0.155625 & 0.153190 \end{array} \right)$$