Mathematics: 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?

Calculus 2 if we have ∫ (sec² b / tan² b) sec b

Ответ:

jaqsway34

11.01.2022

Mathematics

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

$\dfrac{u^2-1+1}{u^2-1}=\dfrac{u^2-1}{u^2-1}+\dfrac1{u^2-1}=1+\dfrac1{u^2-1}$

(provided that $u^2-1\neq0$ , or $u\neq\pm1$ )

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Suppose you had a slightly more complex integrand, like

$\dfrac{u^3}{u^2-1}=u+\dfrac u{u^2-1}$

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 .) 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

$\dfrac{u^3}{u^2-1}=\boxed{u}_{\,q}+\dfrac{\boxed{u}_{\,r}}{u^2-1}$

(where and denote "quotient" and "remainder")

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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?

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