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)

2

Answer 2

$\int \frac{5\,tan\, x }{tan \,x - 2} dx = \int \frac{5\,tan\, x }{sin \,x - 2\,cos\,x} dx = \int \frac{\left(sin \,x - 2\,cos\,x\right)+2\left(cos\,x + 2\,sin\,x\right)}{\left(sin \,x - 2\,cos\,x\right)}dx$