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

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

Vedclass · Accepted Answer

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

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(C) We have $I = \int \frac{\sin \frac{5x}{2}}{\sin \frac{x}{2}} dx$. Using the identity $\sin(A+B) = \sin A \cos B + \cos A \sin B$,we write $\frac{5x}{2} = 2x + \frac{x}{2}$. $I = \int \frac{\sin(2x + \frac{x}{2})}{\sin \frac{x}{2}} dx = \int \frac{\sin 2x \cos \frac{x}{2} + \cos 2x \sin \frac{x}{2}}{\sin \frac{x}{2}} dx$. $I = \int (\sin 2x \cot \frac{x}{2} + \cos 2x) dx$. Alternatively,using the sum-to-product identity or expansion: $\frac{\sin \frac{5x}{2}}{\sin \frac{x}{2}} = \frac{\sin(2x + \frac{x}{2})}{\sin \frac{x}{2}} = \frac{\sin 2x \cos \frac{x}{2} + \cos 2x \sin \frac{x}{2}}{\sin \frac{x}{2}} = \cos 2x + \sin 2x \cot \frac{x}{2}$. Using $\sin 2x = 2 \sin x \cos x$ and $\cot \frac{x}{2} = \frac{1+\cos x}{\sin x}$: $\sin 2x \cot \frac{x}{2} = 2 \sin x \cos x \cdot \frac{1+\cos x}{\sin x} = 2 \cos x(1+\cos x) = 2 \cos x + 2 \cos^2 x$. Since $2 \cos^2 x = 1 + \cos 2x$,we have: $\cos 2x + 2 \cos x + 1 + \cos 2x = 1 + 2 \cos x + 2 \cos 2x$. Integrating: $\int (1 + 2 \cos x + 2 \cos 2x) dx = x + 2 \sin x + \sin 2x + C$.

$\int \frac{\sin \frac{5x}{2}}{\sin \frac{x}{2}} dx = $ (where $C$ is a constant of integration.)

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