If int frac{5 tan x}{tan x-2} dx=ax+b log |si | vedclass

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(B) 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 \co

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$3$

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(B) 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 $5 \sin x = A(\sin x - 2 \cos x) + B \frac{d}{dx}(\sin x - 2 \cos x)$. $5 \sin x = A(\sin x - 2 \cos x) + B(\cos x + 2 \sin x)$. $5 \sin x = (A + 2B) \sin x + (B - 2A) \cos x$. Comparing coefficients: $A + 2B = 5$ and $B - 2A = 0 \implies B = 2A$. Substituting $B = 2A$ into the first equation: $A + 2(2A) = 5 \implies 5A = 5 \implies A = 1$. Then $B = 2(1) = 2$. So,$I = \int \frac{1(\sin x - 2 \cos x) + 2(\cos x + 2 \sin x)}{\sin x - 2 \cos x} dx$. $I = \int 1 dx + 2 \int \frac{\cos x + 2 \sin x}{\sin x - 2 \cos x} dx$. $I = x + 2 \log |\sin x - 2 \cos x| + c$. Comparing with $ax + b \log |\sin x - 2 \cos x| + c$,we get $a = 1$ and $b = 2$. Therefore,$a + b = 1 + 2 = 3$.

If $\int \frac{5 \tan x}{\tan x-2} dx=ax+b \log |\sin x-2 \cos x|+c$,then $a+b=$

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