I have this integral $$ \int_{0}^{\pi/2} \frac{e^{a \cos^2x}}{b^2\cos^2x + c^2\sin^2x}dx.$$
I know that
$$\int_{0}^{\pi/2} \frac{dx}{b^2\cos^2x + c^2\sin^2x}=\frac{\pi}{2bc},$$ and $$\int_{0}^{\pi/2} e^{a \cos^2x}dx=\frac{\pi}{2}e^{a/2} I_n\left(0,\frac{a}{2}\right).$$
Using these integrals is it possible to solve the first integral?
Since $$ \int_{0}^{\pi/2} \cos(2mx) e^{a\cos^2(x)}\,dx =\frac{\pi}{2}e^{a/2} I_m(a/2)\tag{1}$$ it is enough to find the Fourier cosine series of $\frac{1}{b^2\cos^2 x+c^2 \sin^2 x}$ over $\left(-\frac{\pi}{2},\frac{\pi}{2}\right)$. $$\frac{1}{b^2\cos^2 x+c^2 \sin^2 x}=\frac{1}{bc}+\frac{2}{bc}\sum_{m\geq 1}\left(\frac{c-b}{c+b}\right)^m \cos(2mx) \tag{2}$$ immediately leads to $$\int_{0}^{\pi/2}\frac{e^{a\cos^2(x)}}{b^2\cos^2 x+c^2 \sin^2 x}\,dx = \frac{\pi}{2bc}+\frac{\pi e^{a/2}}{bc}\sum_{m\geq 1}\left(\frac{c-b}{c+b}\right)^m I_m(a/2).\tag{3}$$