
Confusion...
$\displaystyle \int \frac{sec^2(x)}{1tan^2(x)}dx$
$\displaystyle u=Tan(x)$
$\displaystyle du=sec^2(x)dx$
$\displaystyle \int \frac{1}{1u^2}du$
Now, I'm really tempted to use arcsin, but the denominator isn't a radical, so how do I deal with 1/1u^2 ?

Re: Confusion...
$\displaystyle \dfrac{1}{1u^2}=\dfrac{1}{2}\left(\dfrac{1}{1+u}+\dfrac{1}{1u}\right)$

Re: Confusion...
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.

Re: Confusion...
it's not "difference of squares" it's "partial fractions".
$\displaystyle \frac{1}{1u^2} = \left(\frac{1}{1+u}\right)\left(\frac{1}{1u}\right)$
suppose we want to write this as:
$\displaystyle \frac{A}{1+u} + \frac{B}{1u}$.
then:
$\displaystyle A(1  u) + B(1 + u) = 1$
that is:
$\displaystyle A  Au + B + Bu = 1 = 1 + 0u$.
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.

Re: Confusion...
$\displaystyle \dfrac{1}{1+u}+\dfrac{1}{1u}=\dfrac{2}{1u^2}\iff \dfrac{1}{2}\left(\dfrac{1}{1+u}+\dfrac{1}{1u}\right)=\dfrac{1}{1u^2}$
Edit: Deveno's post is a more detailed explanation, ignore mine!