int frac{sin frac{5 x}{2}}{sin frac{x}{2}} d | vedclass

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(B) We have $I = \int \frac{\sin \frac{5x}{2}}{\sin \frac{x}{2}} dx$. Using the identity $2 \sin A \cos B = \sin(A+B) + \sin(A-B)$,we multiply the num

Vedclass · Accepted Answer

$x+2 \sin x+2 \sin 2 x+C$

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(B) We have $I = \int \frac{\sin \frac{5x}{2}}{\sin \frac{x}{2}} dx$. Using the identity $2 \sin A \cos B = \sin(A+B) + \sin(A-B)$,we multiply the numerator and denominator by $2 \cos \frac{x}{2}$: $I = \int \frac{2 \sin \frac{5x}{2} \cos \frac{x}{2}}{2 \sin \frac{x}{2} \cos \frac{x}{2}} dx = \int \frac{\sin(3x) + \sin(2x)}{\sin x} dx$. Using the formulas $\sin 3x = 3 \sin x - 4 \sin^3 x$ and $\sin 2x = 2 \sin x \cos x$: $I = \int \frac{3 \sin x - 4 \sin^3 x + 2 \sin x \cos x}{\sin x} dx = \int (3 - 4 \sin^2 x + 2 \cos x) dx$. Using $4 \sin^2 x = 2(1 - \cos 2x)$: $I = \int (3 - 2(1 - \cos 2x) + 2 \cos x) dx = \int (3 - 2 + 2 \cos 2x + 2 \cos x) dx$. $I = \int (1 + 2 \cos 2x + 2 \cos x) dx = x + \sin 2x + 2 \sin x + C$.

$\int \frac{\sin \frac{5 x}{2}}{\sin \frac{x}{2}} d x$ is

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