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

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(D) We are given the integral $I = \int \frac{5	an x}{	an x - 2} dx$.First,express $	an x$ as $\frac{\sin x}{\cos x}$:$I = \int \frac{5(\sin x / \c

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$2$

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(D) We are given the integral $I = \int \frac{5	an x}{	an x - 2} dx$.First,express $	an x$ as $\frac{\sin x}{\cos x}$:$I = \int \frac{5(\sin x / \cos x)}{(\sin x / \cos x) - 2} dx = \int \frac{5\sin x}{\sin x - 2\cos x} dx$.We want to express the numerator in terms of the denominator $(\sin x - 2\cos x)$ and its derivative $(\cos x + 2\sin x)$.Let $5\sin x = A(\sin x - 2\cos x) + B(\cos x + 2\sin x)$.Equating coefficients of $\sin x$ and $\cos x$:$A + 2B = 5$$-2A + B = 0 \implies B = 2A$.Substituting $B = 2A$ into the first equation: $A + 2(2A) = 5 \implies 5A = 5 \implies A = 1$.Then $B = 2(1) = 2$.So,$5\sin x = 1(\sin x - 2\cos x) + 2(\cos x + 2\sin x)$.Substituting this into the integral:$I = \int \frac{(\sin x - 2\cos x) + 2(\cos x + 2\sin x)}{\sin x - 2\cos x} dx$$I = \int 1 dx + 2 \int \frac{\cos x + 2\sin x}{\sin x - 2\cos x} dx$.Let $t = \sin x - 2\cos x$,then $dt = (\cos x + 2\sin x) dx$.$I = x + 2 \int \frac{1}{t} dt = x + 2 \ln |t| + k = x + 2 \ln |\sin x - 2\cos x| + k$.Comparing this with the given form $x + a \ln |\sin x - 2\cos x| + k$,we get $a = 2$.

If $\int \frac{5\tan x}{\tan x - 2} dx = x + a \ln |\sin x - 2\cos x| + k$,then $a$ is equal to

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