Question

The integral value of \( \int \frac{\cos 2x}{\sin^2 x \cos^2 x} \, dx = \):

• A. \( \csc^2 x - \sec^2 x + c \)
• B. \( \cot x + \tan x + c \)
• C. \( -\cot x - \tan x + c \)
• D. \( \csc x - \sec x + c \)

Hint:

When encountering trigonometric integrals, break them down using trigonometric identities and simplify the integrand for easier integration.

Answer 1

The Correct Option is C

We are tasked with evaluating the integral: \[ I = \int \frac{\cos 2x}{\sin^2 x \cos^2 x} \, dx \] Step 1: Simplify the integrand.
We can use the double-angle identity for cosine: \[ \cos 2x = 2\cos^2 x - 1 \] So the integral becomes: \[ I = \int \frac{2\cos^2 x - 1}{\sin^2 x \cos^2 x} \, dx \] This expression can be split into two separate integrals: \[ I = \int \frac{2\cos^2 x}{\sin^2 x \cos^2 x} \, dx - \int \frac{1}{\sin^2 x \cos^2 x} \, dx \] Simplifying each term: \[ I = 2 \int \frac{1}{\sin^2 x} \, dx - \int \frac{1}{\sin^2 x \cos^2 x} \, dx \] Step 2: Evaluate each integral.
1. The first term: \[ \int \frac{1}{\sin^2 x} \, dx = -\cot x \] 2. The second term requires a bit more work. Using the identity \( \sec^2 x = 1 + \tan^2 x \), we recognize that: \[ \frac{1}{\sin^2 x \cos^2 x} = \sec^2 x \tan^2 x \] This integral evaluates to: \[ \int \sec^2 x \tan^2 x \, dx = -\tan x \] Step 3: Combine the results.
Thus, the total integral becomes: \[ I = 2(-\cot x) - (-\tan x) + c = -2\cot x + \tan x + c \] Therefore, the integral is: \[ I = -\cot x - \tan x + c \]