Question

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

• A. -1
• B. -2
• C. 1
• D. 2

Answer 1

The correct option is D 2

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