5.4. INTEGRATION BY PARTS
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At this point, all that remains to do is evaluate the (simpler) integral
sin(x) · 1 dx. Doing
so, we find
x cos(x) dx = x sin(x) − (− cos(x)) + C = x sin(x) + cos(x) + C.
There are at least two additional important observations to make from Example 5.3. First,
the general technique of Integration by Parts involves trading the problem of integrating
the product of two functions for the problem of integrating the product of two related
functions. In particular, we convert the problem of evaluating
u dv for that of evaluating
v du. This perspective clearly shapes our choice of u and v. In Example 5.3, the original
integral to evaluate was
x cos(x) dx, and through the substitution provided by Integration
by Parts, we were instead able to evaluate
sin(x) · 1 dx. Note that the original function
x was replaced by its derivative, while cos(x) was replaced by its antiderivative. Second,
observe that when we get to the final stage of evaluating the last remaining antiderivative,
it is at this step that we include the integration constant, +C.
Activity 5.10.
Evaluate each of the following indefinite integrals. Check each antiderivative that you
find by differentiating.
(a)
te −t dt
(b)
4x sin(3x) dx
(c)
z sec 2 (z) dz
(d)
x ln(x) dx
⊳
Some Subtleties with Integration by Parts
There are situations where Integration by Parts is not an obvious choice, but the technique
is appropriate nonetheless. One guide to understanding why is the observation that
integration by parts allows us to replace one function in a product with its derivative while
replacing the other with its antiderivative. For instance, consider the problem of evaluating
arctan(x) dx.
Initially, this problem seems ill-suited to Integration by Parts, since there does not appear
to be a product of functions present. But if we note that arctan(x) = arctan(x) · 1, and
realize that we know the derivative of arctan(x) as well as the antiderivative of 1, we
303
At this point, all that remains to do is evaluate the (simpler) integral
sin(x) · 1 dx. Doing
so, we find
x cos(x) dx = x sin(x) − (− cos(x)) + C = x sin(x) + cos(x) + C.
There are at least two additional important observations to make from Example 5.3. First,
the general technique of Integration by Parts involves trading the problem of integrating
the product of two functions for the problem of integrating the product of two related
functions. In particular, we convert the problem of evaluating
u dv for that of evaluating
v du. This perspective clearly shapes our choice of u and v. In Example 5.3, the original
integral to evaluate was
x cos(x) dx, and through the substitution provided by Integration
by Parts, we were instead able to evaluate
sin(x) · 1 dx. Note that the original function
x was replaced by its derivative, while cos(x) was replaced by its antiderivative. Second,
observe that when we get to the final stage of evaluating the last remaining antiderivative,
it is at this step that we include the integration constant, +C.
Activity 5.10.
Evaluate each of the following indefinite integrals. Check each antiderivative that you
find by differentiating.
(a)
te −t dt
(b)
4x sin(3x) dx
(c)
z sec 2 (z) dz
(d)
x ln(x) dx
⊳
Some Subtleties with Integration by Parts
There are situations where Integration by Parts is not an obvious choice, but the technique
is appropriate nonetheless. One guide to understanding why is the observation that
integration by parts allows us to replace one function in a product with its derivative while
replacing the other with its antiderivative. For instance, consider the problem of evaluating
arctan(x) dx.
Initially, this problem seems ill-suited to Integration by Parts, since there does not appear
to be a product of functions present. But if we note that arctan(x) = arctan(x) · 1, and
realize that we know the derivative of arctan(x) as well as the antiderivative of 1, we