Question

$\int \frac{dx}{\sin^2x \cos^2x}$ is equal to

• A. tan x + cot x + C
• B. (tan x + co
• C. tan x - cotx + C
• D. (tan x - c

Answer 1

Answer: (C)

tan x - cot x + C

Answer 2

To evaluate the integral $\int \frac{dx}{\sin^2x \cos^2x}$, we can use trigonometric identities.

We know that $\sin^2x + \cos^2x = 1$. We can substitute this into the numerator of the integrand: $\int \frac{1}{\sin^2x \cos^2x} dx = \int \frac{\sin^2x + \cos^2x}{\sin^2x \cos^2x} dx$

Now, we can split the fraction into two terms: $\int \left( \frac{\sin^2x}{\sin^2x \cos^2x} + \frac{\cos^2x}{\sin^2x \cos^2x} \right) dx$

Simplify each term: $\int \left( \frac{1}{\cos^2x} + \frac{1}{\sin^2x} \right) dx$

Recall the reciprocal trigonometric identities: $\frac{1}{\cos x} = \sec x$ and $\frac{1}{\sin x} = \csc x$. So, $\frac{1}{\cos^2x} = \sec^2x$ and $\frac{1}{\sin^2x} = \csc^2x$. $\int (\sec^2x + \csc^2x) dx$

Now, integrate each term separately using the standard integration formulas: $\int \sec^2x dx = \tan x + C_1$ $\int \csc^2x dx = -\cot x + C_2$

Combining these results, the integral is: $\tan x - \cot x + C$ where $C = C_1 + C_2$ is the constant of integration.