Question

\[\int\frac{\sin 4x - 2}{1 - \cos 4x} e^{2x} \text{ dx}\]

Answer 1

\[\text{We have}, \]
\[I = \int\left( \frac{\sin 4x - 2}{1 - \cos 4x} \right) e^{2x} \text{ dx}\]
\[ = \int\left( \frac{2 \sin 2x \cos 2x - 2}{2 \sin^2 2x} \right) e^{2x} \text{ dx}\]
\[ = \int\left[ \cot \left( 2x \right) - {cosec}^2 \left( 2x \right) \right] e^{2x} \text{ dx}\]
\[\text{ Let e}^{2x} \cot \left( 2x \right) = t\]
\[ \Rightarrow \left[ 2 e^{2x} \cot \left( 2x \right) + e^{2x} \left\{ - {cosec}^2 \left( 2x \right) \right\} \times 2 \right] dx = dt\]
\[ \Rightarrow e^{2x} \left[ \cot 2x - {cosec}^2 \left( 2x \right) \right] dx = \frac{dt}{2}\]
\[ \therefore I = \frac{1}{2}\int dt\]
\[ = \frac{t}{2} + C\]
\[ = \frac{1}{2}\text{  e}^{2x} \text{ cot } \left( \text{ 2x} \right) + C\]