I apologize in advance

Note: I am not convinced that this is the simplest solution possible, but it was the only one I could think of up until the posting of this solution.

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Let $I$ be $\int{\sin{(x)}\,{\cos{(3x)}}\; {\sin{(5x)}}\; {\cos{(7x)}}\; {\sin{(9x)}}\; {\cos{(11x)}}\; \mathrm{d}x}$

$$

By the product-to-sum formulae,

$\sin{x}\cos{3x}=\frac{1}{2}(\sin{4x}-\sin{2x})$ ,

$\sin{5x}\cos{7x}=\frac{1}{2}(\sin{12x}-\sin{2x})$ , and

$\sin{9x}\cos{11x}=\frac{1}{2}(\sin{20x}-\sin{2x})$ .

$$

We therefore have:

$I=\frac{1}{8}\int{(\sin{4x}-\sin{2x})\cdot{(\sin{12x}-\sin{2x})}\cdot{(\sin{20x}-\sin{2x})}\;\mathrm{d}x}$

$=\frac{1}{8}\int{(\sin{4x}\cdot{\sin{12x}}-\sin{4x}\cdot{\sin{2x}}-\sin{2x}\cdot{\sin{12x}}+\sin^{2}{2x})\cdot{(\sin{20x}-\sin{2x})}\;\mathrm{d}x}$

$$

Again by product-to-sum formulae, and after simplification,

$I=\frac{1}{16}\int{(-c(16x)+c(8x)+c(6x)-c(2x)+c(14x)-c(10x)-c(4x)+1)\cdot{(s(20x)-s(2x))}\;\mathrm{d}x}$

where $c(f(x))$ indicates the cosine of $f(x)$ and $s(f(x))$ indicates the sine of $f(x)$ .

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After expanding, and simplifying using yet again the product-to-sum formulae, we finally arrive at

$I=\frac{1}{16}\int{(-\frac{3}{2}s(2x)+\frac{1}{2}s(4x)+\frac{3}{2}s(6x)-s(8x)-s(10x)+\frac{3}{2}s(12x)+s(16x)+s(20x)-\frac{1}{2}s(22x)\;...}$

$...\;-\frac{1}{2}s(24x)+\frac{1}{2}s(26x)+\frac{1}{2}s(28x)-\frac{1}{2}s(30x)+\frac{1}{2}s(34x)-\frac{1}{2}s(36x))\;\mathrm{d}x$

which integrates to give us

$\frac{3\cos(2x)}{64}-\frac{\cos(4x)}{128}-\frac{\cos(6x)}{64}+\frac{\cos(8x)}{128}+\frac{\cos(10x)}{160}-\frac{\cos(12x)}{128}+\frac{\cos(16x)}{256}-\frac{\cos(20x)}{320}+\frac{\cos(22x)}{704}+\frac{\cos(24x)}{768}\;...$

$...\;-\;\frac{\cos(26x)}{832}-\frac{\cos(28x)}{896}+\frac{\cos(30x)}{960}-\frac{\cos(34x)}{1088}+\frac{\cos(36x)}{1152}+C$

$$

Finally, substituting our upper limit $\frac{\pi}{4}$ and lower limit $0$ into the integral, we get $\frac{3797}{24\,504\,480}$ which cannot be further simplified.

$$

Therefore, $a+b=3797+24\,504\,480=\boxed{24\,508\,277}$