MCQ
$\int_{}^{} {\tan (3x - 5)\sec (3x - 5)\;dx = } $
  • A
    $\sec (3x - 5) + c$
  • $\frac{1}{3}\sec (3x - 5) + c$
  • C
    $\tan (3x - 5) + c$
  • D
    None of these

Answer

Correct option: B.
$\frac{1}{3}\sec (3x - 5) + c$
b
(b) Put $t = 3x - 5 \Rightarrow dt = 3dx,$ therefore
$\int_{}^{} {\tan (3x - 5)\,\sec (3x - 5)\,dx} = \frac{1}{3}\int_{}^{} {\tan t.\sec t\,dt} $
$ = \frac{{\sec t}}{3} + c = \frac{{\sec (3x - 5)}}{3} + c.$

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