If int frac{5 tan x}{tan x-2} d x = alpha x + | vedclass

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(A) We are given the integral $I = \int \frac{5 	an x}{	an x - 2} dx = \int \frac{5 \sin x}{\sin x - 2 \cos x} dx$. Let $5 \sin x = \alpha(\sin x -

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$-1$

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(A) We are given the integral $I = \int \frac{5 	an x}{	an x - 2} dx = \int \frac{5 \sin x}{\sin x - 2 \cos x} dx$. Let $5 \sin x = \alpha(\sin x - 2 \cos x) + \beta \frac{d}{dx}(\sin x - 2 \cos x)$. $5 \sin x = \alpha(\sin x - 2 \cos x) + \beta(\cos x + 2 \sin x)$. Comparing the coefficients of $\sin x$ and $\cos x$: For $\sin x$: $5 = \alpha + 2\beta$ $(i)$ For $\cos x$: $0 = -2\alpha + \beta \implies \beta = 2\alpha$ (ii) Substituting (ii) into $(i)$: $5 = \alpha + 2(2\alpha) = 5\alpha \implies \alpha = 1$. Then $\beta = 2(1) = 2$. Thus,$5 \sin x = 1(\sin x - 2 \cos x) + 2(\cos x + 2 \sin x)$. Substituting this into the integral: $I = \int \frac{1(\sin x - 2 \cos x) + 2(\cos x + 2 \sin x)}{\sin x - 2 \cos x} dx$ $I = \int 1 dx + 2 \int \frac{\cos x + 2 \sin x}{\sin x - 2 \cos x} dx$ $I = x + 2 \log |\sin x - 2 \cos x| + \gamma$. Comparing this with the given form $\alpha x + \beta \log |\sin x - 2 \cos x| + \gamma$,we get $\alpha = 1$ and $\beta = 2$. Therefore,$\alpha - \beta = 1 - 2 = -1$.

If $\int \frac{5 \tan x}{\tan x-2} d x = \alpha x + \beta \log |\sin x - 2 \cos x| + \gamma$,then $\alpha - \beta =$

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