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

Question

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

Aicharya · Accepted Answer

We have, $ \int \frac{5 	an x}{	an x-2} d x=x+a \log |\sin x-2 \cos x|+K $ On differentiating, we get $ \begin{aligned} & \frac{5 	an x}{	an x-2}=1+\frac{a(\cos x+2 \sin x)}{\sin x-2 \cos x} \ \Rightarrow \quad \frac{5 	an x}{	an x-2} & =1+\frac{a(1+2 	an x)}{	an x-2} \ \Rightarrow \quad & \frac{5 	an x}{	an x-2}=\frac{	an x-2+a+2 a 	an x}{	an x-2} \ \Rightarrow \quad 5 	an x & =(2 a+1) 	an x+a-2 \end{aligned} $ Equating the coefficient of $	an x$ and constant terms, we get $ \begin{aligned} 2 a+1 & =5 \ a-2 & =0 \ 	herefore \quad a & =2 \end{aligned} $

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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