Question

$\int\frac{dx}{sin^{2}x~cos^{2}x}=$

• A. $tan~x+cot~x+c$
• B. $tan~x-cot~x+c$
• C. $tan~x~cot~x+c$
• D. $tan~x-cot~2x+c$

Hint:

$\int sec^{2}x~dx = tan~x$ and $\int cosec^{2}x~dx = -cot~x$.

Answer 1

The Correct Option is B

Step 1: Understanding the Question:
Evaluate an indefinite integral involving trigonometric functions in the denominator.

Step 2: Key Formula or Approach:
Use the identity $\sin^2 x + \cos^2 x = 1$ to split the fraction.

Step 3: Detailed Explanation:
\[ I = \int \frac{1}{\sin^2 x \cos^2 x} dx \] Substitute $1 = \sin^2 x + \cos^2 x$: \[ I = \int \frac{\sin^2 x + \cos^2 x}{\sin^2 x \cos^2 x} dx \] \[ I = \int (\frac{\sin^2 x}{\sin^2 x \cos^2 x} + \frac{\cos^2 x}{\sin^2 x \cos^2 x}) dx \] \[ I = \int (\frac{1}{\cos^2 x} + \frac{1}{\sin^2 x}) dx \] \[ I = \int (\sec^2 x + \csc^2 x) dx \] Integrating term by term: \[ I = \tan x - \cot x + c \]

Step 4: Final Answer:
The integral is $\tan x - \cot x + c$.