Just Integral :)

$\begin{aligned} I & = \int_0^\infty \frac {\cos (4x)}{169x^2+ 1} dx & \small \blue{\text{Since the integrand is even.}} \\ & = \frac 12 \int_{-\infty}^\infty \frac {\cos (4x)}{169x^2+ 1} dx \\ & = \frac 1{338} \int_{-\infty}^\infty \frac {\cos (4x)}{x^2+ \frac 1{169}} dx & \small \blue{\text{Consider upper semicircle in the complex}} \\ & = \frac 1{338} \int_C \frac {e^{4ix}}{z^2+ \frac 1{169}} dx & \small \blue{\text{plain with }R \to \infty \text{ as contour}} \\ & = \frac 1{338} \int_C \frac {e^{4iz}}{\left(z-\frac i{13}\right)\left(z+\frac i{13}\right)} dx & \small \blue{\text{There is one singularity at }z = \frac i{13}} \\ & = \frac {2\pi i}{338} \cdot \frac {e^{-\frac 4{13}}}{\frac {2i}{13}} & \small \blue{\text{By Cauchy integral formula}} \\ & = \frac \pi{26 e^\frac 4{13}} \end{aligned}$

Therefore $d-c+b-a = 13-4+26-1 = \boxed{34}$ .

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References:

• Contour integration
• Cauchy integral formula

The value of the integral is $\dfrac{π}{26e^{\dfrac{4}{13}}}$ . So $a=1,b=26,c=4,d=13$ and $d-c+b-a=13-4+26-1=\boxed {34}$