Question

$\int \frac{1}{\sin x \cos x}\, dx$ is equal to:

• A. $\log|\tan x| + C$
• B. $\log|\sin 2x| + C$
• C. $\log|\sec x| + C$
• D. $\log|\cos x| + C$
• E. $\log|\sin x| + C$

Hint:

An alternative quick way is to multiply the numerator and denominator by 2 to get $\int \frac{2}{\sin 2x} \, dx = 2 \int \text{cosec } 2x \, dx$. This leads to $\log|\tan x| + C$ via standard integration formulas.

Answer 1

The Correct Option is A

Concept: There are several ways to approach this, but multiplying and dividing by a factor to create a standard derivative or using a double-angle identity is most efficient.

Step 1: Convert the integrand into a more useful form.
Divide the numerator and denominator by $\cos^2 x$: \[ \int \frac{1 / \cos^2 x}{(\sin x \cos x) / \cos^2 x} \, dx = \int \frac{\sec^2 x}{\tan x} \, dx \]

Step 2: Apply substitution.
Let $u = \tan x$, then $du = \sec^2 x \, dx$. The integral becomes: \[ \int \frac{1}{u} \, du = \log|u| + C \]

Step 3: Substitute back.
\[ \log|\tan x| + C \]