Q36 / 90 · JEE Main 2012 (Offline) · Mathematics

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

  • A
    $-1$
  • B
    $-2$
  • C
    $1$
  • Correct answer
    D
    $2$

Solution

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

Hence, the correct answer is option: 'D'.

Relevant Questions