$\int \frac{s i n 3 x}{c o s 4 x c o s x} d x$ is equal to

- \frac{1}{4} log | cos 4_{X} | - log | cos x | + c

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- \frac{1}{4} log | cos 4 x | + log | cos x | + c

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- \frac{1}{4} log | sin 4 x | + log | cos 4 x | + c

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\frac{1}{4} log | sin 4 x | - log | cos 4 x | + c

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Solution

The correct option is A $- \frac{1}{4} log | cos 4 x | + log | cos x | + c$
$\int \frac{s i n 3 x}{c o s 4 x c o s x} d x$

$= \int \frac{s i n (4 x - x)}{c o s 4 x \cdot c o s x} d x$

$= \int \frac{[s i n (4 x) c o s x - c o s (4 x) s i n x]}{(c o s 4 x) (c o s x)} d x$

$= \int t a n 4 x d x - \int t a n x d x$

$= \frac{1}{4} l o g | s e c 4 x | - l o g | s e c x | + c$

$= \frac{- 1}{4} l o g | c o s 4 x | + l o g | c o s x | + c$

$\int \frac{s i n 3 x}{c o s 4 x c o s x} d x = \frac{- 1}{4} l o g | c o s 4 x | + l o g | c o s x | + c$

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