Question

\[ \int \frac{dx}{\sin^2x\cos^2x}= \]

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

Hint:

When denominator contains \(\sin^2x\cos^2x\), use \(1=\sin^2x+\cos^2x\) to split the expression.

Answer 1

The Correct Option is B

Concept: We use the trigonometric identity: \[ \sin^2x+\cos^2x=1 \] Also, \[ \sec^2x=\frac{1}{\cos^2x}, \qquad \cosec^2x=\frac{1}{\sin^2x} \]

Step 1: Given integral is: \[ \int \frac{dx}{\sin^2x\cos^2x} \]

Step 2: Write numerator \(1\) as: \[ 1=\sin^2x+\cos^2x \] So, \[ \frac{1}{\sin^2x\cos^2x} = \frac{\sin^2x+\cos^2x}{\sin^2x\cos^2x} \]

Step 3: Split the fraction. \[ \frac{\sin^2x}{\sin^2x\cos^2x} + \frac{\cos^2x}{\sin^2x\cos^2x} \] \[ = \frac{1}{\cos^2x} + \frac{1}{\sin^2x} \] \[ = \sec^2x+\cosec^2x \]

Step 4: Now integrate. \[ \int \frac{dx}{\sin^2x\cos^2x} = \int (\sec^2x+\cosec^2x)\,dx \] \[ = \int \sec^2x\,dx+\int \cosec^2x\,dx \]

Step 5: Use standard integration formulas. \[ \int \sec^2x\,dx=\tan x \] and \[ \int \cosec^2x\,dx=-\cot x \] Therefore, \[ \int \frac{dx}{\sin^2x\cos^2x} = \tan x-\cot x+c \] Hence, \[ \boxed{\tan x-\cot x+c} \]