If int frac{cos (13 x)-cos (14 x)}{1+2 cos (9 | vedclass

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(A) Given,$I = \int \frac{\cos (13x) - \cos (14x)}{1 + 2 \cos (9x)} dx$. Multiply the numerator and denominator by $\sin(9x/2)$: $I = \int \frac{(\cos

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$4^5$

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(A) Given,$I = \int \frac{\cos (13x) - \cos (14x)}{1 + 2 \cos (9x)} dx$. Multiply the numerator and denominator by $\sin(9x/2)$: $I = \int \frac{(\cos 13x - \cos 14x) \sin(9x/2)}{(1 + 2 \cos 9x) \sin(9x/2)} dx$. Using the identity $1 + 2 \cos(9x) = \frac{\sin(27x/2)}{\sin(9x/2)}$,we get: $I = \int \frac{(\cos 13x - \cos 14x) \sin(9x/2)}{\sin(27x/2)} dx$. Using the product-to-sum formula $\cos A - \cos B = 2 \sin(\frac{A+B}{2}) \sin(\frac{B-A}{2})$: $\cos 13x - \cos 14x = 2 \sin(\frac{27x}{2}) \sin(x/2)$. Substituting this into the integral: $I = \int \frac{2 \sin(27x/2) \sin(x/2) \sin(9x/2)}{\sin(27x/2)} dx = \int 2 \sin(9x/2) \sin(x/2) dx$. Using $2 \sin A \sin B = \cos(A-B) - \cos(A+B)$: $I = \int (\cos(4x) - \cos(5x)) dx = \frac{\sin 4x}{4} - \frac{\sin 5x}{5} + C$. Comparing with the given form,$a = 4$ and $b = 5$. Therefore,$a^b = 4^5$.

If $\int \frac{\cos (13 x)-\cos (14 x)}{1+2 \cos (9 x)} d x=\frac{\sin (4 x)}{a}-\frac{\sin (5 x)}{b}+c$,then $a^b=$

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