Evaluate: $\int e^{x} (\frac{\sin 4 x - 4}{1 - \cos 4 x}) dx$

Solution

$Let I = \int e^{x} (\frac{\sin 4 x - 4}{1 - \cos 4 x}) dx = \int e^{x} (\frac{\sin 2 (2 x) - 4}{1 - \cos 2 (2 x)}) dx = \int e^{x} (\frac{2 \sin 2 x \cos 2 x - 4}{2 \sin^{2} (2 x)}) dx [Using, \sin 2 x = 2 sinx . cosx and 2 \sin^{2} x = 1 - \cos (2 x)] = \int e^{x} (\frac{2 (\sin (2 x) \cos (2 x) - 2)}{2 \sin^{2} (2 x)}) dx = \int e^{x} (\frac{\sin (2 x) \cos (2 x)}{\sin^{2} (2 x)} - \frac{2}{\sin^{2} (2 x)}) dx = \int e^{x} (\cot (2 x) - 2 {cosec}^{2} (2 x)) dx Now, let f (x) = \cot (2 x) then f (x) = - 2 {cosec}^{2} (2 x) I = \int e^{x} (f (x) + f' (x)) dx$

$So, I = e^{x} f (x) + C = e^{x} \cot (2 x) + C, Where C is a constant therefore, \int e^{x} (\frac{\sin 4 x - 4}{1 - \cos 4 x}) dx = e^{x} \cot (2 x) + C$

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Evaluate: $\int e^{x} (\frac{\sin 4 x - 4}{1 - \cos 4 x}) dx$