_ ^ ^2 x ;dx ^2 x + 4 = - Vidyadip

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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