If int e^{alpha x}left(frac{1-beta sin x}{1-c | vedclass

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(B) Given, $\int e^{\alpha x}\left(\frac{1-\beta \sin x}{1-\cos x}\right) d x=-e^x \cot \frac{x}{2}+c$. By differentiating both sides with respect to

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(B) Given, $\int e^{\alpha x}\left(\frac{1-\beta \sin x}{1-\cos x}\right) d x=-e^x \cot \frac{x}{2}+c$. By differentiating both sides with respect to $x$, we get: $e^{\alpha x}\left(\frac{1-\beta \sin x}{1-\cos x}\right) = \frac{d}{dx} \left(-e^x \cot \frac{x}{2}\right) + 0$. Using the product rule, $\frac{d}{dx}(uv) = u'v + uv'$: $e^{\alpha x}\left(\frac{1-\beta \sin x}{1-\cos x}\right) = -\left[e^x \cot \frac{x}{2} + e^x \left(-\frac{1}{2} \csc^2 \frac{x}{2}\right)\right]$. $e^{\alpha x}\left(\frac{1-\beta \sin x}{1-\cos x}\right) = -e^x \left[\frac{\cos(x/2)}{\sin(x/2)} - \frac{1}{2 \sin^2(x/2)}\right]$. $e^{\alpha x}\left(\frac{1-\beta \sin x}{1-\cos x}\right) = -e^x \left[\frac{2 \sin(x/2) \cos(x/2) - 1}{2 \sin^2(x/2)}\right]$. Using $2 \sin(x/2) \cos(x/2) = \sin x$ and $2 \sin^2(x/2) = 1 - \cos x$: $e^{\alpha x}\left(\frac{1-\beta \sin x}{1-\cos x}\right) = -e^x \left[\frac{\sin x - 1}{1 - \cos x}\right] = e^x \left(\frac{1 - \sin x}{1 - \cos x}\right)$. Comparing both sides, we get $\alpha = 1$ and $\beta = 1$. Therefore, $\frac{\alpha^2 + \beta^2}{2 \alpha \beta} = \frac{1^2 + 1^2}{2(1)(1)} = \frac{2}{2} = 1$.

If $\int e^{\alpha x}\left(\frac{1-\beta \sin x}{1-\cos x}\right) d x=-e^x \cot \frac{x}{2}+c$, then $\frac{\alpha^2+\beta^2}{2 \alpha \beta}=$

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