If $\int \frac{5 \tan x}{\tan x-2} d x=x+a \log |\sin x-2 \cos x|+c$, then $a$ (Where $\mathrm{c}$ is…

If $\int \frac{5 \tan x}{\tan x-2} d x=x+a \log |\sin x-2 \cos x|+c$, then $a$ (Where $\mathrm{c}$ is constant of integration)
  1. 1
  2. -2
  3. -1
  4. 2

Solution

Let $I=\int \frac{5 \tan x}{\tan x-2} d x$ $I=\int \frac{5 \sin x}{\sin x-2 \cos x} d x$ Here $\frac{d}{d x}(\sin x-2 \cos x)=\cos x+2 \sin x$ $\begin{aligned} \therefore \mathrm{I} & =\int \frac{(2 \sin x+2 \sin x+\sin x)+(2 \cos x-2 \cos x)}{\sin x-2 \cos x} d x \\ & =\int \frac{(2 \sin x+\cos x)+(2 \sin x+\cos x)+(\sin x-2 \cos x)}{\sin x-2 \cos x} d x \\ & =\int \frac{2(2 \sin x+\cos x)+(\sin x-2 \cos x)}{\sin x-2 \cos x} d x \\ & =\int d x+2 \int \frac{2 \sin x+\cos x}{\sin x-2 \cos x} d x \\ & =x+2 \log |\sin x-2 \cos x|+c \end{aligned}$ From given data, $a=2$

Asked in: MHT CET 2021 (21 Sep Shift 2)

Practice more Indefinite Integration questions on Aicharya