Question

Calculus 2

If we have
∫ (sec² B / tan² B) sec B tan B dB
And u = secB
Du = secBtanBdB
And we get ∫(u^2)/(u^2-1) du
How do we go from there to :
∫ ( 1 + (1/(u^2-1) du
Can somebody explain to me fully why it became that last expression?

Answer 1

A simple way to see what was done is to add and subtract -1 from the numerator:

`(u^2-1+1)/(u^2-1)=(u^2-1)/(u^2-1)+\frac1{u^2-1}=1+\frac1{u^2-1}`

(provided that
`u^2-1\\eq0`, or
`u\\eq\pm1`)

###

Suppose you had a slightly more complex integrand, like

`(u^3)/(u^2-1)=u+\frac u{u^2-1}`

How do we know that? (Assume we don't already know the previous result, so that it's not just a matter of multiplying both sides by
`u`.) Simple polynomial division:

`u^3=\boxed{u}_(\,q)\cdot u^2`, and
`u(u^2-1)=u^3-u`. Subtracting this from
`u^3` gives a remainder of
`u^3-(u^3-u)=\boxed{u}_(\,r)`, so

`(u^3)/(u^2-1)=\boxed{u}_(\,q)+\frac{\boxed{u}_(\,r)}{u^2-1}`

(where
`q` and
`r` denote "quotient" and "remainder")