Substitution Method
Math 132
Stewart §4.5
Reversing the Chain Rule. As we have seen from the
R b Second Fundamental
Theorem (§4.3), the easiest way to evaluate an integral a f (x) dx is to find an
R
Rb
antiderivative, the indefinite integral f (x) dx = F (x) + C, so that a f (x) dx =
F (b)−F (a). Building on §3.9, we will learn several methods to find antiderivatives
which reverse our methods of differentiation, in this case the Chain Rule.
For example, let us find the antiderivative:
Z
x cos(x2 ) dx .
That is, for what function will the Derivative Rules produce x cos(x2 )? We notice
an inside function g(x) = x2 , and a factor x which is very close to the derivative
g 0 (x) = 2x. In fact, we can get the exact derivative of the inside function if we
multiply the factors by 21 and 2:
x cos(x2 ) =
1
2
cos(x2 ) · (2x) .
This is just the kind of derivative function produced by the Chain Rule:
??
f (g(x))0 = f 0 (g(x)) · g 0 (x) = f 0 (x2 ) · (2x) =
1
2
cos(x2 ) · (2x) .
We still need to find the outside function f . To remind us of the original inside
function, we write f (u), where the new variable u represents u = g(x) = x2 . We
must get f 0 (u) = 12 cos(u), an easy antiderivative:
Z
1
1
2 cos(u) du = f (u) + C = 2 sin(u) + C .
Now we restore the original inside function to get our final answer:
Z
2
1
1
1
2 cos(u) du = 2 sin(u) + C = 2 sin(x ) + C .
The Chain Rule in Leibnitz notation (§2.5) reverses and checks the above
computation. Writing y = 21 sin(u) and u = x2 :
dy du
d
dy
=
·
=
dx
du dx
du
=
1
2
cos(u) · (2x) =
1
2
1
2
d 2
sin(u) ·
x
dx
cos(x2 ) · (2x) = x cos(x2 ) .
Substitution Method
R
1. Given an antiderivative h(x) dx, try to find an inside function g(x) such
that g 0 (x) is a factor of the integrand:
h(x) = f (g(x)) · g 0 (x).
This will often involve multiplying and dividing by a constant to get the
exact derivative g 0 (x). After factoring out g 0 (x), sometimes the remaining
factor needs to be manipulated to write it as a function of u = g(x).
du
dx = g 0 (x) dx, write:
2. Using the symbolic notation u = g(x), du =
dx
Z
Z
Z
0
h(x) dx =
f (g(x)) · g (x) dx =
f (u) du ,
and find the antiderivative
R
f (u) du = F (u) + C by whatever method.
3. Restore the original inside function:
Z
Z
h(x) dx =
f (u) du = F (u) + C = F (g(x)) + C .
Examples
R
√
• (3x+4) 3x+4 dx. The inside function is clearly u = 3x+4, du = 3 dx, so:
Z
Z
√
√
1
(3x+4) 3x+4 dx =
3 (3x+4) 3x+4 · 3 dx
Z
=
Z
•
√
1
3u
u du =
1
3
Z
u3/2 du =
1 2 5/2
3 5u
+C =
5/2
2
15 (3x+4)
+ C.
√
√
√
x 3x+4 dx. Again u = 3x+4, so 3x+4 becomes u, but we must still
express the remaining factor x in terms of u. We solve u = 3x+4 to obtain
x = 31 u − 43 : that is, x = 31 (3x+4) − 43 :
Z
Z
Z
√
√
√
4
4
1 1
1 1
x 3x+4 dx =
3 ( 3 (3x+4)− 3 ) 3x+4 · 3 dx =
3 ( 3 u− 3 ) u du
Z
=
Z
•
1 3/2 4 1/2
− 9 u du
9u
=
1 2 5/2 4 2 3/2
− 9 3 u +C
9 5u
=
5/2
2
8
− 27
(3x+4)3/2 +C.
45 (3x+4)
√
√
sec2 ( x)
√
dx . We take u = x = x1/2 , du = 12 x−1/2 dx = 2√1 x dx
x
√
Z
Z
√
sec2 ( x)
1
√
dx =
2 sec2 ( x) · √ dx
x
2 x
Z
√
=
sec2 (u) du = tan(u) + C = tan( x) + C.
Here we use the trig integrals from §3.9.
Z
sin(x)
•
dx. We cannot take the inside function u = sin(x), because
(1 + cos(x))2
its derivative cos(x) is not a factor of the integrand. We could take u =
cos(x), but the best choice is u = 1 + cos(x), du = − sin(x) dx:
Z
Z
sin(x)
1
dx
=
−
· (− sin(x)) dx
(1 + cos(x))2
(1 + cos(x))2
Z
1
1
1
du =
+C =
+ C.
= −
2
u
u
1 + cos(x)
Z
•
√
√
√
√
1− x
p √ dx. Take u = 1+ x, du = 2√1 x , so
x = u−1, 1− x = 2−u.
1+ x
√
√
Z
Z
√
1− x
1− x
1
p √ dx =
p √ (2 x) · √ dx
2 x
1+ x
1+ x
Z
2−u
√ 2(u−1) du = 2
u
=
Z
= 2
u3/2 − 3u1/2 + 2u−1/2 du =
=
√
4
5 (1+
Z
u2 −3u+2
√
du
u
4 5/2
5u
− 43 u3/2 + 8u1/2 + C
√
√
x)5/2 − 43 (1+ x)3/2 + 8(1+ x)1/2 + C.
1
√
Whew! Here we did not have the derivative factor du
dx = 2 x already present:
we had to multipy and divide by it to get du,
R then express the remaining
factors in terms of u. By luck, the resulting du was do-able.
Z
•
sec2 (x) tan(x) dx. Here we could take u = tan(x), du = sec2 (x) dx:
Z
Z
2
sec (x) tan(x) dx =
Z
=
u du =
1 2
2u
tan(x) · sec2 (x) dx
+C =
1
2
tan2 (x) + C.
Alternatively, use the inside function z = sec(x), dz = tan(x) sec(x) dx:
Z
Z
sec2 (x) tan(x) dx =
sec(x) · tan(x) sec(x) dx
Z
=
z dz =
1 2
2z
+C =
1
2
sec2 (x) + C.
Thus 21 tan2 (x) and 12 sec2 (x) are two different antiderivatives, so what about
the Antiderivative Uniqueness Theorem (§3.9)? In fact, the identity tan2 (x)+
1 = sec2 (x) implies:
=
1
2
sec2 (x) .
These give the same antiderivate family:
1
2
tan2 (x) + C =
1
2
tan2 (x) +
1
2
1
2
sec2 (x) + C 0 !
Substitution for definite integrals. We have, for u = g(x):
Z b
Z g(b)
0
f (g(x)) g (x) dx =
f (u) du .
a
example:
g(a)
R3
x(1+x2 )5 dx. Taking u = 1+x2 , du = 2x dx:
Z 4
Z 4
Z 1+42
2 5
2 5
1
1 5
x(1+x ) dx =
2 (1+x ) · 2x dx =
2 u du
2
1+32
3
3
1 6 u=17
= 12
u u=10 =
6
1
12 10
−
6
1
12 17 .
Integral Symmetry
Theorem: If f (x) is an odd function, meaning f (−x) =
Ra
−f (x), then −a f (x) dx = 0.
Proof. By the Integral Splitting Rule (§4.2), we have:
Z a
Z 0
Z a
f (x) dx =
f (x) dx +
f (x) dx .
−a
−a
0
Substituting u = −x, du = (−1) dx in the first term:
Z 0
Z 0
Z 0
f (−x) · (−1) dx
−f (x) · (−1) dx =
f (x) dx =
−a
−a
−a
Z
0
Z
f (u) du = −
=
a
a
Z
f (u) du = −
0
a
f (x) dx .
0
The last equality holds because the variable
and
Ra
R a is merely suggestive,
R 0 of integration
can be changed arbitrarily. Therefore −a f (x) dx + 0 f (x) dx = − 0 f (x) dx +
Ra
0 f (x) dx = 0, as desired.
Rπ
example: Evaluate the definite integral −π x cos(x) dx. Here substitution will
not work, and it is difficult to find an antiderivative. But since (−x) cos(−x) =
−(x cos(x)), the Theorem tells us the integral must be zero.
Geometrically, the integral is the signed area between the graph and the x-axis:
Since the function f (x) = x cos(x) is odd, the graph has rotational symmetry
around the origin, and each negative area below the x-axis cancels a positive area
above the x-axis.