Substitution
The Chain Rule is used to differentiate a composite function. For example,
d
sin(x2 ) = cos(x2 )2x.
dx
Restating as an antiderivative
Z
cos(x2 )2x dx = sin(x2 ) + C.
In general,
Z
d
f (g(x)) = f 0 (g(x))g 0 (x) =⇒
dx
f 0 (g(x))g 0 (x) dx = f (g(x)) + C.
This leads to the procedure which we refer to as ‘u-Substitution’.
Substitution: If u = g(x) is a differentiable function and f (x) is continuous, then
Z
Z
f (g(x))g 0 (x) dx = f (u) du.
Observations:
• We introduce a “new variable” u to represent the “inside function” g(x).
• For u = g(x),
du
dx
= g 0 (x), so that du = u0 (x)dx and
Z
Z
f (g(x))g 0 (x) dx = f (u) du.
• Substitution allows us to rewrite a complicated integral in terms of a simpler (easier to evaluate)
integral.
• We go from an integral involving only the variable x to an integral involving only the variable u.
Z
p
Example 1. Evaluate
x2 1 + x3 dx.
Intuition: The function 1 + x3 is “inside” another function and its derivative (up to a constant multiple)
appears outside the function.
Let u = 1 + x3 , so that du = 3x2 dx and 13 du = x2 dx.
Z
x2
p
1 + x3 dx =
Z p
1 + x3 x2 dx
√ 1
u du
3
Z
1 √
=
u du
3
1 2 3/2
=
u +C
33
3/2
2
=
+ C.
1 + x3
9
Z
=
Observations:
1
Calculus II Resources
Integration Techniques
• The final step when using a u-Substituion to solve an indefinite integral is to rewrite (i.e. resubstitute)
in terms of the original variable x.
• We can differentiate to check our solution. The Chain Rule yields
3/2
1/2 2
32
d 2
+C =
3x
1 + x3
1 + x3
dx 9
29
p
= x2 1 + x3 .
See solution video
Z
Example 2. Evaluate
1
dx.
x ln(x)
Intuition: The function ln(x) is “inside” another function (i.e. in the denominator) and its derivative
appears outside the function.
Let u = ln(x), so that du = x1 dx.
Z
1
dx =
x ln(x)
Z
1 1
dx
ln(x) x
Z
1
=
du
u
= ln |u| + C
= ln | ln(x)| + C
See solution video
Z
Example 3. Evaluate
√
x 2x − 1 dx.
√
Intuition: The integrand x 2x − 1 has the general form we look for when performing a u-Substitution. In
order to choose u “inside” another function we let u = 2x − 1.
Let u = 2x − 1, so that du = 2dx and 12 du = dx. In order to rewrite our integral so that it only involves
the variable u, we somehow need to express the leftover x in terms of u. To do so, we will always use the
original u-substitution. In this case, u = 2x − 1 implies x = 21 (u + 1).
Z
√
x 2x − 1 dx =
√ 1
1
(u + 1) u du
2
2
Z
1
3/2
1/2
=
u +u
du
4
1 2 5/2 2 3/2
=
u + u
+C
4 5
3
1
1
=
(2x − 1)5/2 + (2x − 1)3/2 + C
10
6
Z
Observations:
• In order to express any leftover x’s in terms of u, we use the original u-substitution.
• Good choices for u include something “inside” another function or things you cannot easily integrate
(see Example 2).
See solution video
2
Calculus II Resources
Integration Techniques
Z
Example 4. Evaluate
0
π/2
sin(x)
dx.
1 + cos2 (x)
Observation: When applying a u-Substitution to a definite integral, the limits of integration will change.
In general, if u = g(x), then
Z x=b
Z u=g(b)
0
f (g(x))g (x) dx =
f (u) du.
x=a
u=g(a)
Let u = cos(x), so that du = − sin(x)dx and note that x = 0 → u = cos(0) = 1, and x =
0.
Z
x=π/2
x=0
sin(x)
dx =
1 + cos2 (x)
Z
u=0
u=1
1
π
2
→ u = cos
π
2
=
−1
du
1 + u2
Z
1
du
1
+
u2
0
1
= arctan(u)
=
0
= arctan(1) − arctan(0)
π
= .
4
Observations: When using a u-Substitution to solve a definite integral you MUST change the limits of
integration.
See solution video
Summary:
• Z
The method of u-Substitution is an integration technique used to evaluate integrals of the form
f (g(x))g 0 (x) dx. The substitution u = g(x) implies du = u0 (x)dx so that
Z
Z
0
f (g(x))g (x) dx =
f (u) du.
• To choose u we look for an “inside” function g(x) with g 0 (x) appearing in the integrand.
• A u-Substitution expresses a complicated integral in terms of a simpler (hopefully easier to evaluate)
integral.
• When applying a u-Substitution to an indefinite integral, the final step is to resubstitute in terms of
the original variable. See Examples 1, 2, and 3.
• When applying a u-Substitution to a definite integral, you MUST change the limits of integration. See
Example 4.
See Substitution overview video
Practice Problems: Evaluate the following Integrals
√
Z
(1)
Z
(2)
e x
√ dx
x
√
Z
√
x2 x + 1 dx
Z
1
ex
dx
1 + e2x
2
x
dx
(x + 1)3
(4)
4 + 3x dx
(5)
0
Z
(3)
Z
tan(x) dx
(6)
0
3
Calculus II Resources
Integration Techniques
See Solutions
4