Integrate the function ^2 x ^2 x + 4 - Vidyadip

Question

Integrate the function $\frac{{{{\sec }^2}x}}{{\sqrt {{{\tan }^2}x + 4} }}$

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Answer

Let $I = \int {\frac{{{{\sec }^2}x}}{{\sqrt {{{\tan }^2}x + 4} }}dx} $ ...(i)
Putting tan x = t
$ \Rightarrow {\sec ^2}x = \frac{{dt}}{{dx}}$
$\Rightarrow {\sec ^2}xdx = dt$
$\therefore $ From eq. (i),
$I = \int {\frac{{dt}}{{\sqrt {{t^2} + 4} }}} $
$= \int {\frac{1}{{\sqrt {{t^2} + {{\left( 2 \right)}^2}} }}dt} $
$ = \log \left| {t + \sqrt {{t^2} + {{\left( 2 \right)}^2}} } \right| + c$
$= \log \left| {\tan x + \sqrt {{{\tan }^2}x + 4} } \right| + c$

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