Given, \(I=\int \frac{\cos 13 x-\cos 14 x}{1+2 \cos 9 x} d x\)
Let,
\(\begin{aligned}
f(x) & =\frac{\cos 13 x-\cos 14 x}{1+2 \cos 9 x} \\
& =\frac{\sin 9 x(\cos 13 x-\cos 14 x)}{\sin 9 x+\sin 18 x} \\
& =\frac{\sin 9 x-2 \sin \frac{27}{2} x \cdot \sin \left(-\frac{x}{2}\right)}{2 \sin \frac{27 x}{2} \cdot \cos \left(-\frac{9 x}{2}\right)} \\
& =\frac{\sin 9 x \cdot \sin \frac{x}{2}}{\cos \left(\frac{9 x}{2}\right)}=2 \sin \frac{9 x}{2} \cdot \sin \frac{x}{2} \\
& =\cos \left(\frac{9 x}{2}-\frac{x}{2}\right)-\cos \left(\frac{9 x}{2}+\frac{x}{2}\right) \\
& =\cos 4 x-\cos 5 x
\end{aligned}\)
So,
\(\begin{aligned}
I & =\int(\cos 4 x-\cos 5 x) d x \\
& =\frac{\sin 4 x}{4}-\frac{\sin 5 x}{5}+C
\end{aligned}\)
Hence, \(a^b=4^5\)