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int (sin " x")/(cos 3x.cos 2x) dx =...

`int (sin " x")/(cos 3x.cos 2x)` dx =

A

`(1)/(3) " log cos 3x" + (1)/(2) "log cos " 2x + c `

B

`(1)/(3) " log | sec 3 x|" (1)/(2) ` log | sec 2x | + c

C

`(1)/(3) ` log | tan 3x | - `(1)/(2)` log | tan 2x | + c

D

none

Text Solution

Verified by Experts

The correct Answer is:
B
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