_ ^ dx 4 ^3 2x - 3 2x = - Vidyadip

MCQ

$\int_{}^{} {\frac{{dx}}{{4{{\cos }^3}2x - 3\cos 2x}}} = $

A
$\frac{1}{3}\log [\sec 6x + \tan 6x] + c$
✓
$\frac{1}{6}\log [\sec 6x + \tan 6x] + c$
C
$\log [\sec 6x + \tan 6x] + c$
D
None of these

✓

Answer

Correct option: B.

$\frac{1}{6}\log [\sec 6x + \tan 6x] + c$

b
(b)$\int_{}^{} {\frac{{dx}}{{4{{\cos }^3}2x - 3\cos 2x}}} = \int_{}^{} {\frac{{dx}}{{\cos 6x}} = \int_{}^{} {\sec 6xdx} } $
$ = \frac{1}{6}\log \,(\sec 6x + \tan 6x) + c.$

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