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

Question

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

Aicharya · Accepted Answer

$\int \frac{5 	an x}{	an x-2} d x=\int \frac{5 \sin x}{\sin x-2 \cos x} d x$ $\Rightarrow \int\left[\frac{2(\cos x+2 \sin x)+(\sin x-2 \cos x)}{\sin x-2 \cos x}ight] d x$ $=2 \int\left(\frac{\cos x+2 \sin x}{\sin x-2 \cos x}ight) d x+\int d x+k$ $=2 \log |\sin x-2 \cos x|+x+k \quad 	herefore a=2$

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

Solution