MCQ
$\int_{}^{} {{{\tan }^4}x\;dx = } $
  • A
    ${\tan ^3}x - \tan x + x + c$
  • $\frac{1}{3}{\tan ^3}x - \tan x + x + c$
  • C
    $\frac{1}{3}{\tan ^3}x + \tan x + x + c$
  • D
    $\frac{1}{3}{\tan ^3}x + \tan x + 2x + c$

Answer

Correct option: B.
$\frac{1}{3}{\tan ^3}x - \tan x + x + c$
b
(b)$\int_{}^{} {{{\tan }^4}x\,dx} = \int_{}^{} {{{\tan }^2}x({{\sec }^2}x - 1)\,dx} $
$ = \int_{}^{} {{{\tan }^2}x{{\sec }^2}x\,dx} - \int_{}^{} {{{\tan }^2}x\,dx} = \frac{{{{\tan }^3}x}}{3} - \tan x + x + c$.

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