If int frac{5 tan (x)}{tan (x)-2} d x = x + a | vedclass

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(D) We have,$\int \frac{5 	an x}{	an x-2} d x = x + a \log |\sin x - 2 \cos x| + K$. On differentiating both sides with respect to $x$,we get: $\fra

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$2$

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(D) We have,$\int \frac{5 	an x}{	an x-2} d x = x + a \log |\sin x - 2 \cos x| + K$. On differentiating both sides with respect to $x$,we get: $\frac{5 	an x}{	an x-2} = 1 + a \frac{d}{dx} (\log |\sin x - 2 \cos x|) $ $\frac{5 	an x}{	an x-2} = 1 + a \frac{\cos x + 2 \sin x}{\sin x - 2 \cos x} $ Dividing the numerator and denominator of the fraction by $\cos x$: $\frac{5 	an x}{	an x-2} = 1 + a \frac{1 + 2 	an x}{	an x - 2} $ $\frac{5 	an x}{	an x-2} = \frac{	an x - 2 + a + 2a 	an x}{	an x - 2} $ Comparing the coefficients of $	an x$ and the constant terms: $5 = 2a + 1 \Rightarrow 2a = 4 \Rightarrow a = 2$ Also,$a - 2 = 0 \Rightarrow a = 2$. Thus,$a = 2$.

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

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