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Integrate the functions\[\frac{\sec^2x}{\sqrt{\tan^2x+4}}\]

$\begin{array}{1 1}\log\mid\tan x+\sqrt{\tan^2x+4}\mid+c \\ \log\mid\tan x-{\tan^2x+4}\mid+c \\ \log\mid\tan x-\sqrt{\tan^2x+4}\mid+c \\ \log\mid\tan x+\sqrt{\sec^2x+4}\mid+c \end{array} $

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  • $\int\frac{dx}{\sqrt{x^2+a^2}}=\int log\mid x+\sqrt{x^2+a^2}\mid+c.$
Given:$I=\int\frac{\sec^2x}{\sqrt{\tan^2x+4}}dx.$
 
Put tan x=t.
 
On differentiating we get,
 
$\sec^2xdx=dt.$
 
Substituting this we get,
 
$I=\int\frac{dt}{\sqrt {t^2+2^2}}$.
 
On integrating we get,
 
$\;\;\;=log\mid t+\sqrt{t^2+4}\mid+c.$
 
Substituting for t we get,
 
$\;\;\;=log\mid\tan x+\sqrt{\tan^2x+4}\mid+c.$

 

answered Feb 4, 2013 by sreemathi.v
edited Feb 4, 2013 by sreemathi.v
 
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