$\int \frac{c o s 4 x + 1}{c o t x - t a n x} d x$

\frac{1}{8} c o s 4 x + C

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\frac{- 1}{8} c o s 4 x + C

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\frac{1}{4} c o s 2 x + C

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\frac{1}{2} c o s x + C

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Solution

The correct option is B $\frac{- 1}{8} c o s 4 x + C$
$\int (\frac{c o s 4 x + 1}{c o t x - t a n x}) d x = \int (\frac{2 c o s^{2} 2 x - 1 + 1}{(\frac{c o s x}{s i n x}) - (\frac{s i n x}{c o s x})}) d x = \int (\frac{2 c o s^{2} 2 x \times s i n x c o s x}{c o s^{2} x - s i n^{2} x}) d x = \int (\frac{2 c o s^{2} 2 x s i n x c o s x}{c o s 2 x}) d x = \int c o s 2 x \times 2 s i n x c o s x d x = \int c o s 2 x \times s i n 2 x d x = (\frac{1}{2}) \int 2 s i n x 2 x c o s 2 x d x = (\frac{1}{2}) \int s i n 4 x d x = (\frac{- c o s 4 x}{8}) + c$

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