Question

If the integral
$\int \frac{5\,tan\, x }{tan \,x - 2} dx = x + a \ell n |sin x - 2 cos x| + k$, then $a$ is equal to :

• A. -1
• B. -2
• C. 1
• D. 2

Answer 1

Answer: (D)

$\int \frac{5\,tan\, x }{tan \,x - 2} dx = \int \frac{5\,tan\, x }{sin \,x - 2\,cos\,x} dx = \int \frac{\left(sin \,x - 2\,cos\,x\right)+2\left(cos\,x + 2\,sin\,x\right)}{\left(sin \,x - 2\,cos\,x\right)}dx$
$= \int dx +2\int \frac{cos \,x + 2\,sin\,x}{sin \,x - 2\,cos\,x} dx = x + 2ln |\left(sin\, x - 2 \,cos\, x\right)| + k$
$\Rightarrow \quad a = 2$