Question

\[\int e^x \left( \frac{\sin 4x - 4}{1 - \cos 4x} \right) dx\]

Answer 1

\[\text{ Let I } = \int e^x \left[ \frac{\sin4x - 4}{1 - \cos4x} \right]dx\]

\[ = \int e^x \left[ \frac{2\sin2x \cos2x}{2 \sin^2 (2x)} - \frac{4}{2 \sin^2 2x} \right]dx\]

\[ = \int e^x \left[ \cot(2x) - \text{ 2
}{cosec}^2 (2x) \right]dx\]

\[\text{ Here,} f(x) = \text
{ cot } (2x)\]

\[ \Rightarrow f'(x) = - \text{ 2 }{cosec}^2 (2x)\]

\[\text{ Put e}^x f(x) = t\]

\[\text{ let e }^x \text{ cot }( 2x) = t\]

\[\text{ Diff  both  sides w . r . t x}\]

\[ e^x \text{ cot (2x) } + e^x \times \left[ - \text{ 2  } {cosec}^\text{ 2 }(2x) \right] = \frac{dt}{dx}\]

\[ \Rightarrow e^x \left[ \cot(2x) -\text{  2 } {cosec}^\text{ 2 }(2x) \right]dx = dt\]

\[ \therefore \int e^x \left[ \cot 2x - \text{ 2 }{cosec}^2 \text{ 2x } \right]dx = \int dt\]

\[ \Rightarrow I = t + C\]

\[ = e^x \cot\left( \text{ 2x } \right) + C\]