5.4. INTEGRATION BY PARTS
305
Using Integration by Parts Multiple Times
We have seen that the technique of Integration by Parts is well suited to integrating the
product of basic functions, and that it allows us to essentially trade a given integrand for a
new one where one function in the product is replaced by its derivative, while the other is
replaced by its antiderivative. The main goal in this trade of
u dv for
v du is to have
the new integral not be more challenging to evaluate than the original one. At times, it
turns out that it can be necessary to apply Integration by Parts more than once in order to
ultimately evaluate a given indefinite integral.
For example, if we consider
t 2 e t dt and let u = t 2 and dv = e t dt, then it follows that
du = 2t dt and v = e t , thus
t
2 e
t dt = t
2 e
t −
2te
t dt.
The integral on the righthand side is simpler to evaluate than the one on the left, but it
still requires Integration by Parts. Now letting u = 2t and dv = e t dt, we have du = 2 dt
and v = e t , so that
t
2 e
t dt = t
2 e
t −
2te
t −
2e
t dt
.
Note the key role of the parentheses, as it is essential to distribute the minus sign to the
entire value of the integral
2te t dt. The final integral on the right in the most recent
equation is a basic one; evaluating that integral and distributing the minus sign, we find
t
2 e
t dt = t
2 e
t − 2te
t + 2e
t + C.
Of course, situations are possible where even more than two applications of Integration
by Parts may be necessary. For instance, in the preceding example, it is apparent that if
the integrand was t 3 e t instead, we would have to use Integration by Parts three times.
Next, we consider the slightly different scenario presented by the definite integral
e t cos(t) dt. Here, we can choose to let u be either e t or cos(t); we pick u = cos(t), and
thus dv = e t dt. With du = − sin(t) dt and v = e t , Integration by Parts tells us that
e
t cos(t) dt = e
t cos(t) −
e
t (− sin(t)) dt,
or equivalently that
e
t cos(t) dt = e
t cos(t) +
e
t sin(t) dt
(5.9)
Observe that the integral on the right in Equation (5.9),
e t sin(t) dt, while not being
more complicated than the original integral we want to evaluate, it is essentially identical
305
Using Integration by Parts Multiple Times
We have seen that the technique of Integration by Parts is well suited to integrating the
product of basic functions, and that it allows us to essentially trade a given integrand for a
new one where one function in the product is replaced by its derivative, while the other is
replaced by its antiderivative. The main goal in this trade of
u dv for
v du is to have
the new integral not be more challenging to evaluate than the original one. At times, it
turns out that it can be necessary to apply Integration by Parts more than once in order to
ultimately evaluate a given indefinite integral.
For example, if we consider
t 2 e t dt and let u = t 2 and dv = e t dt, then it follows that
du = 2t dt and v = e t , thus
t
2 e
t dt = t
2 e
t −
2te
t dt.
The integral on the righthand side is simpler to evaluate than the one on the left, but it
still requires Integration by Parts. Now letting u = 2t and dv = e t dt, we have du = 2 dt
and v = e t , so that
t
2 e
t dt = t
2 e
t −
2te
t −
2e
t dt
.
Note the key role of the parentheses, as it is essential to distribute the minus sign to the
entire value of the integral
2te t dt. The final integral on the right in the most recent
equation is a basic one; evaluating that integral and distributing the minus sign, we find
t
2 e
t dt = t
2 e
t − 2te
t + 2e
t + C.
Of course, situations are possible where even more than two applications of Integration
by Parts may be necessary. For instance, in the preceding example, it is apparent that if
the integrand was t 3 e t instead, we would have to use Integration by Parts three times.
Next, we consider the slightly different scenario presented by the definite integral
e t cos(t) dt. Here, we can choose to let u be either e t or cos(t); we pick u = cos(t), and
thus dv = e t dt. With du = − sin(t) dt and v = e t , Integration by Parts tells us that
e
t cos(t) dt = e
t cos(t) −
e
t (− sin(t)) dt,
or equivalently that
e
t cos(t) dt = e
t cos(t) +
e
t sin(t) dt
(5.9)
Observe that the integral on the right in Equation (5.9),
e t sin(t) dt, while not being
more complicated than the original integral we want to evaluate, it is essentially identical