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STD 12 - 7.1 inderfinite integral — Mathematics STD 12 Science — Question

CBSE BoardEnglish MediumSTD 12 ScienceMathematicsSTD 12 - 7.1 inderfinite integral1 Mark
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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