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5.3. INTEGRATION BY SUBSTITUTION
Reversing the Chain Rule: u-substitution
Of course, a natural question arises from our recent work: what happens when the inner
function is not a linear function? For example, can we find antiderivatives of such functions
as
g(x) = xe
x 2 and h(x) = e
x 2 ?
It is important to explicitly remember that differentiation and antidifferentiation are
essentially inverse processes; that they are not quite inverse processes is due to the +C
that arises when antidifferentiating. This close relationship enables us to take any known
derivative rule and translate it to a corresponding rule for an indefinite integral. For
example, since
d
dx
x
5
= 5x
4
,
we can equivalently write
5x
4 dx = x
5 + C.
Recall that the Chain Rule states that
d
dx
[ f (g(x))] = f
′ (g(x)) · g
′ (x).
Restating this relationship in terms of an indefinite integral,
f
′ (g(x))g
′ (x) dx = f (g(x)) + C.
(5.5)
Hence, Equation (5.5) tells us that if we can take a given function and view its algebraic
structure as f ′ (g(x))g ′ (x) for some appropriate choices of f and g, then we can antidifferentiate the function by reversing the Chain Rule. It is especially notable that both
g(x) and g ′ (x) appear in the form of f ′ (g(x))g ′ (x); we will sometimes say that we seek to
identify a function-derivative pair when trying to apply the rule in Equation (5.5).
In the situation where we can identify a function-derivative pair, we will introduce a
new variable u to represent the function g(x). Observing that with u = g(x), it follows in
Leibniz notation that
du
dx = g ′ (x), so that in terms of differentials 4 , du = g ′ (x) dx. Now
converting the indefinite integral of interest to a new one in terms of u, we have
f
′ (g(x))g
′ (x) dx =
f
′ (u) du.
Provided that f ′ is an elementary function whose antiderivative is known, we can now
4 If we recall from the definition of the derivative that du
dx ≈ △u
△x and use the fact that du
dx = g ′ (x), then
we see that g ′ (x) ≈ △u
△x . Solving for △u, △u ≈ g ′ (x)△x. It is this last relationship that, when expressed in
“differential” notation enables us to write du = g ′ (x) dx in the change of variable formula.
5.3. INTEGRATION BY SUBSTITUTION
Reversing the Chain Rule: u-substitution
Of course, a natural question arises from our recent work: what happens when the inner
function is not a linear function? For example, can we find antiderivatives of such functions
as
g(x) = xe
x 2 and h(x) = e
x 2 ?
It is important to explicitly remember that differentiation and antidifferentiation are
essentially inverse processes; that they are not quite inverse processes is due to the +C
that arises when antidifferentiating. This close relationship enables us to take any known
derivative rule and translate it to a corresponding rule for an indefinite integral. For
example, since
d
dx
x
5
= 5x
4
,
we can equivalently write
5x
4 dx = x
5 + C.
Recall that the Chain Rule states that
d
dx
[ f (g(x))] = f
′ (g(x)) · g
′ (x).
Restating this relationship in terms of an indefinite integral,
f
′ (g(x))g
′ (x) dx = f (g(x)) + C.
(5.5)
Hence, Equation (5.5) tells us that if we can take a given function and view its algebraic
structure as f ′ (g(x))g ′ (x) for some appropriate choices of f and g, then we can antidifferentiate the function by reversing the Chain Rule. It is especially notable that both
g(x) and g ′ (x) appear in the form of f ′ (g(x))g ′ (x); we will sometimes say that we seek to
identify a function-derivative pair when trying to apply the rule in Equation (5.5).
In the situation where we can identify a function-derivative pair, we will introduce a
new variable u to represent the function g(x). Observing that with u = g(x), it follows in
Leibniz notation that
du
dx = g ′ (x), so that in terms of differentials 4 , du = g ′ (x) dx. Now
converting the indefinite integral of interest to a new one in terms of u, we have
f
′ (g(x))g
′ (x) dx =
f
′ (u) du.
Provided that f ′ is an elementary function whose antiderivative is known, we can now
4 If we recall from the definition of the derivative that du
dx ≈ △u
△x and use the fact that du
dx = g ′ (x), then
we see that g ′ (x) ≈ △u
△x . Solving for △u, △u ≈ g ′ (x)△x. It is this last relationship that, when expressed in
“differential” notation enables us to write du = g ′ (x) dx in the change of variable formula.