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If the integral $\int \frac{5 \tan x}{\tan x-2} d x=x+a \ln |\sin x-2 \cos x|+k$, then $a$ is equal to
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$2$
$2$
$\int \frac{5 \tan x}{\tan x-2} d x=\int \frac{5 \sin x}{\sin x-2 \cos x} d x$
$\Rightarrow \int\left[\frac{2(\cos x+2 \sin x)+(\sin x-2 \cos x)}{\sin x-2 \cos x}\right] d x$
$=2 \int\left(\frac{\cos x+2 \sin x}{\sin x-2 \cos x}\right) d x+\int d x+k$
$=2 \log |\sin x-2 \cos x|+x+k \quad \therefore a=2$
$\Rightarrow \int\left[\frac{2(\cos x+2 \sin x)+(\sin x-2 \cos x)}{\sin x-2 \cos x}\right] d x$
$=2 \int\left(\frac{\cos x+2 \sin x}{\sin x-2 \cos x}\right) d x+\int d x+k$
$=2 \log |\sin x-2 \cos x|+x+k \quad \therefore a=2$
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