untouchablecuz said:
I know how to integrate by substitution, but I don't really understand why it works? How do we break up du/dx? Why does the method work? I've always thought that du RELIES on dx etc etc. Any help would be appreciated.
Integration by substitution is basically a method of SIMPLIFYING the integral.
A simple example:
∫ 2x(1 + x²)² dx = (1 + x²)³ / 3 + c
because it is obvious that when you differentiate , (1 + x²)³ a 2x must pop out because of the chain rule and the 3 comes out in front. However, since there is no 3 in the integrand, we have to divide the primitive by 3 to maintain equality.
So one way to rewrite this is if
f(x) = (1 + x²) with
f'(x) = 2x, then
∫
2x(
1 + x²)² dx =∫
f'(x) [
f(x)]² dx = [f(x)]³ / 3 + c
Notice that if we differentiate [f(x)]³ / 3 we get 3[f(x)]² f'(x) / 3 = [f(x)]² f'(x) which is the original integrand.
Integration by substitution follows that very same principle but instead of f(x), it is commonly written as u.
u = (1 + x²) {this is
f(x)}
du/dx = 2x {this is
f'(x)}
So instead of ∫
f'(x)[
f(x)]² dx we write it as:
∫
2x(
1 + x²)² dx = ∫
(du/dx) u² dx
= ∫ u² du
= u³ / 3 + c
= (1 + x²)³ / 3 + c
