Question

Find the value of \(\int\frac{dx}{\,sin^2x\,cos^2x}\) = ?

Answer 1

We are tasked with evaluating the integral: \[ I = \int \frac{dx}{\sin^2 x \cos^2 x} \]

Step 1: Rewrite the integral: The given integral is: \[ I = \int \frac{1}{\sin^2 x \cos^2 x} \, dx \] This expression can be rewritten as: \[ I = \int \frac{\sin^2 x + \cos^2 x}{\sin^2 x \cos^2 x} \, dx \] Since \( \sin^2 x + \cos^2 x = 1 \), we can simplify this to: \[ I = \int \frac{\sin^2 x}{\sin^2 x \cos^2 x} \, dx + \int \frac{\cos^2 x}{\sin^2 x \cos^2 x} \, dx \]

Step 2: Simplify each integral: Now, simplify the individual integrals: \[ I = \int \frac{1}{\cos^2 x} \, dx + \int \frac{1}{\sin^2 x} \, dx \] These integrals can be recognized as standard trigonometric integrals. The first integral is the integral of \( \sec^2 x \), and the second is the integral of \( \csc^2 x \): \[ I = \int \sec^2 x \, dx + \int \csc^2 x \, dx \]

Step 3: Integrate: The integral of \( \sec^2 x \) is \( \tan x \), and the integral of \( \csc^2 x \) is \( -\cot x \). Thus: \[ I = \tan x - \cot x + C \]

Final Answer: Therefore, the integral is: \[ \int \frac{dx}{\sin^2 x \cos^2 x} = \tan x - \cot x + C \]