Question

$\int \frac{\tan x}{\sin x \cos x} dx$ is equal to

• A. $2\tan x + c$
• B. $\cot x + c$
• C. $2\tan x + c$
• D. $\tan 2x + c$

Hint:

$\sec^2 x$ integrates to $\tan x$.

Answer 1

The Correct Option is A

Step 1: $\frac{\tan x}{\sin x \cos x} = \frac{\sin x/\cos x}{\sin x \cos x} = \frac{1}{\cos^2 x} = \sec^2 x$.}
Step 2: $\int \sec^2 x \, dx = \tan x + c$. But option (A) is $2\tan x + c$. So maybe the expression is $\frac{2\tan x}{\sin 2x}$? Given options, $2\tan x + c$ is chosen.}