Question:medium

If $\int \frac{\sin^{3}x + \cos^{3}x}{\sin^{2}x \cos^{2}x} dx = A\sec x + B\csc x + c$ then $(A, B)$ are}

Show Hint

Always look to split complex fractions with a single term in the denominator.
  • $(1, -1)$
  • $(-1, -1)$
  • $(1, 1)$
  • $(-1, 1)$
Show Solution

The Correct Option is A

Solution and Explanation

Step 1: Understanding the Concept:
To solve this integral, split the numerator and divide each term by the denominator. This transforms complex trigonometric products into standard trigonometric functions that are easy to integrate.
Step 2: Key Formula or Approach:
1. Simplify terms: \(\frac{\sin^3 x}{\sin^2 x \cos^2 x} = \tan x \sec x\) and \(\frac{\cos^3 x}{\sin^2 x \cos^2 x} = \cot x \csc x\).
2. Standard integrals: \(\int \sec x \tan x \, dx = \sec x\) and \(\int \csc x \cot x \, dx = -\csc x\).
Step 3: Detailed Explanation:
Split the integral: \[ \int \frac{\sin^3 x}{\sin^2 x \cos^2 x} \, dx + \int \frac{\cos^3 x}{\sin^2 x \cos^2 x} \, dx \] \[ = \int \frac{\sin x}{\cos^2 x} \, dx + \int \frac{\cos x}{\sin^2 x} \, dx \] \[ = \int \sec x \tan x \, dx + \int \csc x \cot x \, dx \] Using standard integration formulas: \[ = \sec x + (-\csc x) + c = 1 \sec x - 1 \csc x + c \] Comparing with \(A \sec x + B \csc x + c\), we find \(A = 1\) and \(B = -1\).
Step 4: Final Answer:
The values are \((A, B) = (1, -1)\).
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