MCQ
$\int_{}^{} {\frac{{{{\sec }^2}x\;dx}}{{\sqrt {{{\tan }^2}x + 4} }} = } $
  • $\log \left[ {\tan x + \sqrt {{{\tan }^2}x + 4} } \right] + c$
  • B
    $\frac{1}{2}\log \left[ {\tan x + \sqrt {{{\tan }^2}x + 4} } \right] + c$
  • C
    $\log \left[ {\frac{1}{2}\tan x + \frac{1}{2}\sqrt {{{\tan }^2}x + 4} } \right] + c$
  • D
    None of these

Answer

Correct option: A.
$\log \left[ {\tan x + \sqrt {{{\tan }^2}x + 4} } \right] + c$
a
(a) Put $t = \tan x \Rightarrow dt = {\sec ^2}x\,dx,$ then
$\int_{}^{} {\frac{{{{\sec }^2}x\,dx}}{{\sqrt {{{\tan }^2}x + 4} }}} = \int_{}^{} {\frac{1}{{\sqrt {{t^2} + {2^2}} }}} \,dt$
$ = \log [\tan x + \sqrt {{{\tan }^2}x + 4} ] + c.$

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