Question

The integral \( \int \frac{dx}{\sin^2 x \cos^2 x} \) is equal to

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

Hint:

When dealing with integrals involving powers of \( \sin x \) and \( \cos x \) in the denominator, it is often useful to use the identity \( 1 = \sin^2 x + \cos^2 x \) to split the fraction. Alternatively, you can use the identity \( \sin 2x = 2 \sin x \cos x \) to rewrite the denominator as \( \left( \frac{\sin 2x}{2} \right)^2 = \frac{\sin^2 2x}{4} \). Then the integral becomes \( \int \frac{4}{\sin^2 2x} dx = 4 \int \csc^2 2x dx \). Integrating \( \csc^2 ax \) gives \( -\frac{1}{a} \cot ax \). So, \( 4 \int \csc^2 2x dx = 4 \left( -\frac{1}{2} \cot 2x \right) + C = -2 \cot 2x + C \). You can show that \( -2 \cot 2x \) is equivalent to \( \tan x - \cot x \) using the double angle formula for cotangent: \( \cot 2x = \frac{\cot^2 x - 1}{2 \cot x} \). $$-2 \cot 2x = -2 \left( \frac{\cot^2 x - 1}{2 \cot x} \right) = -\frac{\cot^2 x - 1}{\cot x} = -\cot x + \frac{1}{\cot x} = \tan x - \cot x$$

Answer 1

The Correct Option is C

We need to evaluate the integral \( \int \frac{dx}{\sin^2 x \cos^2 x} \). We can rewrite the integrand using the identity \( \sin^2 x + \cos^2 x = 1 \): $$\frac{1}{\sin^2 x \cos^2 x} = \frac{\sin^2 x + \cos^2 x}{\sin^2 x \cos^2 x}$$ $$= \frac{\sin^2 x}{\sin^2 x \cos^2 x} + \frac{\cos^2 x}{\sin^2 x \cos^2 x}$$ $$= \frac{1}{\cos^2 x} + \frac{1}{\sin^2 x}$$ $$= \sec^2 x + \csc^2 x$$ Now, we can integrate term by term: $$\int (\sec^2 x + \csc^2 x) dx = \int \sec^2 x dx + \int \csc^2 x dx$$ We know that the integral of \( \sec^2 x \) is \( \tan x \) and the integral of \( \csc^2 x \) is \( -\cot x \). Therefore, $$\int \sec^2 x dx + \int \csc^2 x dx = \tan x - \cot x + C$$ Thus, \( \int \frac{dx}{\sin^2 x \cos^2 x} = \tan x - \cot x + C \).