Confusion...

**Nervous** · January 1st 2013, 07:04 PM

$\int \frac{sec^2(x)}{1-tan^2(x)}dx$
$u=Tan(x)$
$du=sec^2(x)dx$
$\int \frac{1}{1-u^2}du$

Now, I'm really tempted to use arcsin, but the denominator isn't a radical, so how do I deal with 1/1-u^2 ?

**LordoftheFlies** · January 1st 2013, 07:20 PM

$\dfrac{1}{1-u^2}=\dfrac{1}{2}\left(\dfrac{1}{1+u}+\dfrac{1}{1-u}\right)$

**Nervous** · January 2nd 2013, 04:18 AM

Where does the 1/2 come from? I see the difference of squares part, I guess I'm just confused about the splitting the fraction part.

**Deveno** · January 2nd 2013, 04:34 AM

it's not "difference of squares" it's "partial fractions".

$\frac{1}{1-u^2} = \left(\frac{1}{1+u}\right)\left(\frac{1}{1-u}\right)$

suppose we want to write this as:

$\frac{A}{1+u} + \frac{B}{1-u}$ .

then:

$A(1 - u) + B(1 + u) = 1$

that is:

$A - Au + B + Bu = 1 = 1 + 0u$ .

this means that:

A + B = 1
B - A = 0

we can solve this for A and B:

2B = 1 --> B = 1/2 --> A = 1/2. thus the 2 goes on the bottom.

**LordoftheFlies** · January 2nd 2013, 04:35 AM

$\dfrac{1}{1+u}+\dfrac{1}{1-u}=\dfrac{2}{1-u^2}\iff \dfrac{1}{2}\left(\dfrac{1}{1+u}+\dfrac{1}{1-u}\right)=\dfrac{1}{1-u^2}$

Edit: Deveno's post is a more detailed explanation, ignore mine!

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