How do you use substitution to integrate $(secxtanxdx)/(9+4sec^2x)$ ?

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Answer 1 · 2015-07-26T06:29:12

The problem is solved below.

Let, $Sec x = t$

$implies Sec x tan x * dx = dt$ (Upon differentiating)
Thus, the integral becomes,

$int dt/(9 + 4t^2)$

$= int dt/(3^2 + (2t)^2)$

Now, substituting $2t = z$
We get, $dt = dz/2$

Thus, the integral further becomes,

$1/2 int dz/(3^2 + z^2)$

$= 1/2*1/3 arctan (z/3) + C$
$= 1/6 arctan ((2Sec x)/3) + C$

I guess I'm right.

How do you use substitution to integrate (secxtanxdx)/(9+4sec^2x)(secxtanxdx)/(9+4sec^2x)?