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MHT CET · Maths · Indefinite Integration

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. A 1
  2. B -2
  3. C -1
  4. D 2
Verified Solution

Answer & Solution

Correct Answer

(D) 2

Step-by-step Solution

Detailed explanation

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 \(\frac{d}{d x}(\sin x-2 \cos x)=\cos x+2 \sin x\)
\(\therefore \mathrm{I} =\int \frac{(2 \sin x+2 \sin x+\sin x)+(2 \cos x-2 \cos x)}{\sin x-2 \cos x} d x \)
\( =\int \frac{(2 \sin x+\cos x)+(2 \sin x+\cos x)+(\sin x-2 \cos x)}{\sin x-2 \cos x} d x \)
\( =\int \frac{2(2 \sin x+\cos x)+(\sin x-2 \cos x)}{\sin x-2 \cos x} d x \)
\( =\int d x+2 \int \frac{2 \sin x+\cos x}{\sin x-2 \cos x} d x \)
\( =x+2 \log |\sin x-2 \cos x|+c\)
From given data, \(a=2\)