Question:medium

$\int \text{e}^{2x} \frac{(\sin 2x \cos 2x-1)}{\sin^2 2x} \text{d}x =$

Show Hint

$\int e^{ax} [a f(x) + f'(x)] dx = e^{ax} f(x)$.
Updated On: May 7, 2026
  • $\text{e}^{2x} \cot(2x) + \text{c}$
  • $2\text{e}^{2x} \cot(2x) + \text{c}$
  • $4\text{e}^{2x} \cot(2x) + \text{c}$
  • $-\frac{1}{2} \text{e}^{2x} \cot(2x) + \text{c}$
Show Solution

The Correct Option is D

Solution and Explanation

Step 1: Understanding the Concept:
This integral follows the standard form $\int \text{e}^{ax} [af(x) + f'(x)] \text{d}x = \text{e}^{ax} f(x) + c$. We simplify the trigonometric fraction to match this structure.
Step 2: Key Formula or Approach:
Rewrite the integrand:
\[ \frac{\sin 2x \cos 2x - 1}{\sin^2 2x} = \frac{\sin 2x \cos 2x}{\sin^2 2x} - \frac{1}{\sin^2 2x} = \cot(2x) - \csc^2(2x) \] Our integral is $\int \text{e}^{2x} [\cot(2x) - \csc^2(2x)] \text{d}x$.
Step 3: Detailed Explanation:
Let $f(x) = \frac{1}{2} \cot(2x)$.
Then $f'(x) = \frac{1}{2} [-\csc^2(2x) \cdot 2] = -\csc^2(2x)$.
The given expression in the integral is:
\[ \cot(2x) - \csc^2(2x) = 2 \left[ \frac{1}{2} \cot(2x) \right] + \left[ -\csc^2(2x) \right] = 2f(x) + f'(x) \] Applying the formula $\int \text{e}^{2x} [2f(x) + f'(x)] \text{d}x = \text{e}^{2x} f(x) + c$:
\[ I = \text{e}^{2x} \left( \frac{1}{2} \cot(2x) \right) + c = \frac{1}{2} \text{e}^{2x} \cot(2x) + c \] Step 4: Final Answer:
The result of the integration is $\frac{1}{2} \text{e}^{2x} \cot(2x) + c$.
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