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Summation by Sigma Method

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Question

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

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Solution

$\int e^{x} (\frac{s i n 4 x - 4}{1 - c o s 4 x}) d x = \int e^{x} (\frac{2 s i n 2 x c o s 2 x - 4}{2 {s i n}^{2} 2 x}) d x (∵ s i n 2 x = 2 s i n x c o s x a n d c o s 2 x = 1 - 2 {s i n}^{2} x) = \int e^{x} \frac{2 s i n 2 x c o s 2 x}{2 {s i n}^{2} 2 x} d x - \int e^{x} \frac{4}{2 {s i n}^{2} 2 x} d x = \int e^{x} c o t 2 x d x - 2 \int e^{x} {c o s e c}^{2} 2 x d x (∵ c o s e c x = \frac{1}{s i n x}) i n t e g r a t i n g b y p a r t s; = c o t 2 x . \int e^{x} d x - \int [\frac{d}{d x} c o t 2 x . \int e^{x} d x] d x - 2 \int e^{x} {c o s e c}^{2} 2 x d x = e^{x} . c o t 2 x + 2 \int e^{x} {c o s e c}^{2} 2 x d x - 2 \int e^{x} {c o s e c}^{2} 2 x d x (∵ \frac{d}{d x} c o t x = - {c o s e c}^{2} x) = e^{x} . c o t 2 x + C$

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