MCQ
$\int_{}^{} {{{\sec }^4}x\tan x\;dx = } $ 
  • $\frac{1}{4}{\sec ^4}x + c$
  • B
    $4{\sec ^4}x + c$
  • C
    $\frac{{{{\sec }^3}x}}{3} + c$
  • D
    $3{\sec ^3}x + c$

Answer

Correct option: A.
$\frac{1}{4}{\sec ^4}x + c$
a
(a)$\int_{}^{} {{{\sec }^4}tnx\,dx} = \int_{}^{} {{{\sec }^3}x\sec x\tan x\,dx} $
Put $t = \sec x \Rightarrow dt = \sec x\tan x\,dx,$ then it reduces to
$\int_{}^{} {{t^3}dt} = \frac{{{t^4}}}{4} + c = \frac{1}{4}{\sec ^4}x + c.$

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