∫ d x c o s ( 2 x ) c o s ( 4 x ) is equal to Options: A . &nbsp12√2log|1+√2 sin 2x1−√2 sin 2x|−12(log|sec 2x+tan 2x|)+C B . &nbsp12√2log|1+√2 sin 2x1+√2 sin 2x|−12(log|sec 2x+tan 2x|)+C C . &nbsp1√2log|1+√2 sin 2x1+√2 sin 2x|−12(log|sec 2x−tan 2x|)+C D . &nbspNone of these

Answer: Option A : A ∫ s i n ( 4 x − 2 x ) d x s i n ( 2 x ) c o s ( 2 x ) c o s ( 4 x ) = ∫ s i n ( 4 x ) d x s i n ( 2 x ) c o s ( 4 x ) − ∫ s e c 2 x d x = 2 ∫ c o s 2 x d x c o s 4 x − 1 2 ( l o g | s e c 2 x + t a n 2 x | )

∫dxcos(2x)cos(4x)is equal to

Question

\int \frac{d x}{c o s (2 x) c o s (4 x)}

is equal to

Options:

A . 12√2log|1+√2 sin 2x1−√2 sin 2x|−12(log|sec 2x+tan 2x|)+C

B . 12√2log|1+√2 sin 2x1+√2 sin 2x|−12(log|sec 2x+tan 2x|)+C

C . 1√2log|1+√2 sin 2x1+√2 sin 2x|−12(log|sec 2x−tan 2x|)+C

D . None of these

Answer: Option A
:
A

\int \frac{s i n (4 x - 2 x) d x}{s i n (2 x) c o s (2 x) c o s (4 x)} = \int \frac{s i n (4 x) d x}{s i n (2 x) c o s (4 x)} - \int s e c 2 x d x = 2 \int \frac{c o s 2 x d x}{c o s 4 x} - \frac{1}{2} (l o g | s e c 2 x + t a n 2 x |)

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