If int frac{5 tan x}{tan x-2} , dx = x + a lo | vedclass

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(C) Let $I = \int \frac{5 	an x}{	an x - 2} \, dx = \int \frac{5 \sin x}{\sin x - 2 \cos x} \, dx$. We express the numerator as $5 \sin x = A(\sin x

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$2$

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(C) Let $I = \int \frac{5 	an x}{	an x - 2} \, dx = \int \frac{5 \sin x}{\sin x - 2 \cos x} \, dx$. We express the numerator as $5 \sin x = A(\sin x - 2 \cos x) + B \frac{d}{dx}(\sin x - 2 \cos x)$. $5 \sin x = A(\sin x - 2 \cos x) + B(\cos x + 2 \sin x)$. Equating the coefficients of $\sin x$ and $\cos x$: $A + 2B = 5$ and $-2A + B = 0$. From the second equation,$B = 2A$. Substituting into the first: $A + 2(2A) = 5 \implies 5A = 5 \implies A = 1$. Then $B = 2(1) = 2$. Thus,$I = \int \left( 1 + 2 \frac{\cos x + 2 \sin x}{\sin x - 2 \cos x} ight) \, dx$. $I = \int 1 \, dx + 2 \int \frac{d(\sin x - 2 \cos x)}{\sin x - 2 \cos x} = x + 2 \log |\sin x - 2 \cos x| + c$. Comparing this with $x + a \log |\sin x - 2 \cos x| + c$,we get $a = 2$.

If $\int \frac{5 \tan x}{\tan x-2} \, dx = x + a \log |\sin x - 2 \cos x| + c$ (where $c$ is a constant of integration),then the value of $a$ is

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