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

Answer

Correct option: A.
$\frac{1}{3}{\sec ^3}x - \sec x + c$
a
(a)$\int_{}^{} {\sec x{{\tan }^3}x\;dx = \int_{}^{} {\sec x({{\sec }^2}x - 1)\tan x\;dx} } $
$ = \int_{}^{} {\sec x\tan x{{\sec }^2}x\;dx} - \int_{}^{} {\sec x\tan x\;dx} $
$ = \frac{{{{\sec }^3}x}}{3} - \sec x + c$, (Putting $\sec x = t$ in first part).

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