int frac{cos 7x - cos 8x}{1 + 2 cos 5x} dx = | vedclass

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(B) We have $I = \int \frac{\cos 7x - \cos 8x}{1 + 2 \cos 5x} dx$. Multiply the numerator and denominator by $\sin \frac{5x}{2}$: $I = \int \frac{(\co

Vedclass · Accepted Answer

$\frac{1}{2} \sin 2x - \frac{1}{3} \sin 3x + c$

Vedclass · Answer

(B) We have $I = \int \frac{\cos 7x - \cos 8x}{1 + 2 \cos 5x} dx$. Multiply the numerator and denominator by $\sin \frac{5x}{2}$: $I = \int \frac{(\cos 7x - \cos 8x) \sin \frac{5x}{2}}{(1 + 2 \cos 5x) \sin \frac{5x}{2}} dx$. Using $1 + 2 \cos 5x = 1 + 2(1 - 2 \sin^2 \frac{5x}{2}) = 3 - 4 \sin^2 \frac{5x}{2}$,this does not simplify easily. Alternatively,use the identity $\cos A - \cos B = 2 \sin \frac{A+B}{2} \sin \frac{B-A}{2}$: $I = \int \frac{2 \sin \frac{15x}{2} \sin \frac{x}{2}}{1 + 2 \cos 5x} dx$. Multiply numerator and denominator by $\sin \frac{5x}{2}$: $I = \int \frac{2 \sin \frac{15x}{2} \sin \frac{x}{2} \sin \frac{5x}{2}}{(1 + 2 \cos 5x) \sin \frac{5x}{2}} dx$. Since $\sin 3	heta = 3 \sin 	heta - 4 \sin^3 	heta$,we have $\sin \frac{15x}{2} = 3 \sin \frac{5x}{2} - 4 \sin^3 \frac{5x}{2} = \sin \frac{5x}{2} (3 - 4 \sin^2 \frac{5x}{2}) = \sin \frac{5x}{2} (3 - 2(1 - \cos 5x)) = \sin \frac{5x}{2} (1 + 2 \cos 5x)$. Substituting this back: $I = \int \frac{2 \sin \frac{5x}{2} (1 + 2 \cos 5x) \sin \frac{x}{2}}{(1 + 2 \cos 5x) \sin \frac{5x}{2}} dx = \int 2 \sin \frac{x}{2} \sin \frac{5x}{2} dx$. Using $2 \sin A \sin B = \cos(A-B) - \cos(A+B)$: $I = \int (\cos 2x - \cos 3x) dx = \frac{1}{2} \sin 2x - \frac{1}{3} \sin 3x + c$.

$\int \frac{\cos 7x - \cos 8x}{1 + 2 \cos 5x} dx = $

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