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Find :   \int \frac{sec^2x}{\sqrt{tan^2x+4}}dx

 

 

 

 
 
 
 
 

Answers (1)

Let   I=\int \frac{\sec^2x}{\sqrt{\tan^2x+4}}

Let us assume \tan x=t

Taking derivative with respect to x

\sec^2x\: dx=dt\: \: \: \: -(1)

Substituting (1) in I,

I=\int \frac{dt}{\sqrt{t^2+4}}

    =\int \frac{dt}{\sqrt{t^2+2^2}}

Using formula \int \frac{dx}{\sqrt{x^2+a^2}}=\log |x+\sqrt{x^2+a^2}|+c

I=\log |t+\sqrt{t^2+4}|+c

Substituting t=\tan x

I=\log|\tan x+\sqrt{\tan^2x+4}\, \, |+c

Posted by

Ravindra Pindel

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