Question

\( \int \frac{e^{2x}}{\sin^2 x} \left( 2 \log \csc x + \sin 2x \right) dx = \)

• A. \( -2e^{2x} \log (\csc^2 x) + C \)
• B. \( -2e^{2x} \log (\csc x) + C \)
• C. \( -2e^{2x} \log (\csc x + \sin x) + C \)
• D. \( -2e^{2x} \log (\csc x - \cot x) + C \)

Hint:

Watch for log-trig combinations — they often require product or chain rule thinking.

Answer 1

The Correct Option is B

Let \( I = \int \frac{e^{2x}}{\sin^2 x} (2 \log \csc x + \sin 2x) dx \).
Note that \( \frac{1}{\sin^2 x} = \csc^2 x \), and derivative of \( \log \csc x \) gives terms involving \( \cot x \csc x \), leading toward integration by parts.
The structure leads to simplification where the result becomes:
\[ I = -2e^{2x} \log(\csc x) + C \]