How do you use substitution to integrate $\frac{sec x tan x d x}{9 + 4 {sec}^{2} x}$ $(secxtanxdx)/(9+4sec^2x)$ ?

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How do you use substitution to integrate $\frac{sec x tan x d x}{9 + 4 {sec}^{2} x}$ $(secxtanxdx)/(9+4sec^2x)$ ?

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Answer 1 · 2015-07-26T06:29:12

The problem is solved below.

Explanation:

Let, $sec x = t$ $Sec x = t$

$\Rightarrow sec x tan x \cdot d x = d t$ $implies Sec x tan x * dx = dt$ (Upon differentiating)
Thus, the integral becomes,

$\int \frac{d t}{9 + 4 t^{2}}$ $int dt/(9 + 4t^2)$

$= \int \frac{d t}{3^{2} + {(2 t)}^{2}}$ $= int dt/(3^2 + (2t)^2)$

Now, substituting $2 t = z$ $2t = z$
We get, $d t = \frac{d z}{2}$ $dt = dz/2$

Thus, the integral further becomes,

$\frac{1}{2} \int \frac{d z}{3^{2} + z^{2}}$ $1/2 int dz/(3^2 + z^2)$

$= \frac{1}{2} \cdot \frac{1}{3} arctan (\frac{z}{3}) + C$ $= 1/2*1/3 arctan (z/3) + C$
$= \frac{1}{6} arctan (\frac{2 sec x}{3}) + C$ $= 1/6 arctan ((2Sec x)/3) + C$

I guess I'm right.

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How do you use substitution to integrate $\frac{sec x tan x d x}{9 + 4 {sec}^{2} x}$ $(secxtanxdx)/(9+4sec^2x)$ ?

1 Answer

Explanation:

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How do you use substitution to integrate $\frac{sec x tan x d x}{9 + 4 {sec}^{2} x}$ $(secxtanxdx)/(9+4sec^2x)$ ?