302
5.4. INTEGRATION BY PARTS
Equation (5.7) and solve for it to generate the rule
f (x)g
′ (x) dx = f (x)g(x) −
g(x) f
′ (x) dx.
(5.8)
Often we express Equation (5.8) in terms of the variables u and v, where u = f (x) and
v = g(x). Note that in differential notation, du = f ′ (x) dx and dv = g ′ (x) dx, and thus we
can state the rule for Integration by Parts in its most common form as follows.
u dv = uv −
v du.
To apply Integration by Parts, we look for a product of basic functions that we can
identify as u and dv. If we can antidifferentiate dv to find v, and evaluating
v du is not
more difficult than evaluating
u dv, then this substitution usually proves to be fruitful.
To demonstrate, we consider the following example.
Example 5.3. Evaluate the indefinite integral
x cos(x) dx
using Integration by Parts.
Solution. Whenever we are trying to integrate a product of basic functions through
Integration by Parts, we are presented with a choice for u and dv. In the current problem,
we can either let u = x and dv = cos(x) dx, or let u = cos(x) and dv = x dx. While there
is not a universal rule for how to choose u and dv, a good guideline is this: do so in a way
that
v du is at least as simple as the original problem
u dv.
In this setting, this leads us to choose 6 u = x and dv = cos(x) dx, from which it follows
that du = 1 dx and v = sin(x). With this substitution, the rule for Integration by Parts tells
us that
x cos(x) dx = x sin(x) −
sin(x) · 1 dx.
6 Observe that if we considered the alternate choice, and let u = cos(x) and dv = x dx, then du =
− sin(x) dx and v =
1
2 x 2 , from which we would write
x cos(x) dx =
1
2
x
2 cos(x) −
1
2
x
2 (− sin(x)) dx.
Thus we have replaced the problem of integrating x cos(x) with that of integrating
1
2 x 2 sin(x); the latter is
clearly more complicated, which shows that this alternate choice is not as helpful as the first choice.
5.4. INTEGRATION BY PARTS
Equation (5.7) and solve for it to generate the rule
f (x)g
′ (x) dx = f (x)g(x) −
g(x) f
′ (x) dx.
(5.8)
Often we express Equation (5.8) in terms of the variables u and v, where u = f (x) and
v = g(x). Note that in differential notation, du = f ′ (x) dx and dv = g ′ (x) dx, and thus we
can state the rule for Integration by Parts in its most common form as follows.
u dv = uv −
v du.
To apply Integration by Parts, we look for a product of basic functions that we can
identify as u and dv. If we can antidifferentiate dv to find v, and evaluating
v du is not
more difficult than evaluating
u dv, then this substitution usually proves to be fruitful.
To demonstrate, we consider the following example.
Example 5.3. Evaluate the indefinite integral
x cos(x) dx
using Integration by Parts.
Solution. Whenever we are trying to integrate a product of basic functions through
Integration by Parts, we are presented with a choice for u and dv. In the current problem,
we can either let u = x and dv = cos(x) dx, or let u = cos(x) and dv = x dx. While there
is not a universal rule for how to choose u and dv, a good guideline is this: do so in a way
that
v du is at least as simple as the original problem
u dv.
In this setting, this leads us to choose 6 u = x and dv = cos(x) dx, from which it follows
that du = 1 dx and v = sin(x). With this substitution, the rule for Integration by Parts tells
us that
x cos(x) dx = x sin(x) −
sin(x) · 1 dx.
6 Observe that if we considered the alternate choice, and let u = cos(x) and dv = x dx, then du =
− sin(x) dx and v =
1
2 x 2 , from which we would write
x cos(x) dx =
1
2
x
2 cos(x) −
1
2
x
2 (− sin(x)) dx.
Thus we have replaced the problem of integrating x cos(x) with that of integrating
1
2 x 2 sin(x); the latter is
clearly more complicated, which shows that this alternate choice is not as helpful as the first choice.