5.4. INTEGRATION BY PARTS
307
Evaluating Definite Integrals Using Integration by Parts
Just as we saw with u-substitution in Section 5.3, we can use the technique of Integration
by Parts to evaluate a definite integral. Say, for example, we wish to find the exact value of
π/2
0
t sin(t) dt.
One option is to evaluate the related indefinite integral to find that
t sin(t) dt = −t cos(t)+
sin(t) + C, and then use the resulting antiderivative along with the Fundamental Theorem
of Calculus to find that
π/2
0
t sin(t) dt = (−t cos(t) + sin(t))
π/2
0
=
−
π
2
cos(
π
2
) + sin(
π
2
)
− (−0 cos(0) + sin(0))
= 1.
Alternatively, we can apply Integration by Parts and work with definite integrals
throughout. In this perspective, it is essential to remember to evaluate the product uv over
the given limits of integration. To that end, using the substitution u = t and dv = sin(t) dt,
so that du = dt and v = − cos(t), we write
π/2
0
t sin(t) dt = −t cos(t)
π/2
0
−
π/2
0
(− cos(t)) dt
= −t cos(t)
π/2
0
+ sin(t)
π/2
0
=
−
π
2
cos(
π
2
) + sin(
π
2
)
− (−0 cos(0) + sin(0))
= 1.
As with any substitution technique, it is important to remember the overall goal of the
problem, to use notation carefully and completely, and to think about our end result to
ensure that it makes sense in the context of the question being answered.
When u-substitution and Integration by Parts Fail to Help
As we close this section, it is important to note that both integration techniques we have
discussed apply in relatively limited circumstances. In particular, it is not hard to find
examples of functions for which neither technique produces an antiderivative; indeed,
there are many, many functions that appear elementary but that do not have an elementary