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5.4. INTEGRATION BY PARTS
see the possibility for the substitution u = arctan(x) and dv = 1 dx. We explore this
substitution further in Activity 5.11.
In a related problem, if we consider
t 3 sin(t 2 ) dt, two key observations can be made
about the algebraic structure of the integrand: there is a composite function present
in sin(t 2 ), and there is not an obvious function-derivative pair, as we have t 3 present
(rather than simply t) multiplying sin(t 2 ). This problem exemplifies the situation where
we sometimes use both u-substitution and Integration by Parts in a single problem. If we
write t 3 = t · t 2 and consider the indefinite integral
t · t
2 · sin(t
2 ) dt,
we can use a mix of the two techniques we have recently learned. First, let z = t 2 so
that dz = 2t dt, and thus t dt =
1
2 dz. (We are using the variable z to perform a “zsubstitution” since u will be used subsequently in executing Integration by Parts.) Under
this z-substitution, we now have
t · t
2 · sin(t
2 ) dt =
z · sin(z) ·
1
2
dz.
The remaining integral is a standard one that can be evaluated by parts. This, too, is
explored further in Activity 5.11.
The problems briefly introduced here exemplify that we sometimes must think creatively
in choosing the variables for substitution in Integration by Parts, as well as that it is entirely
possible that we will need to use the technique of substitution for an additional change of
variables within the process of integrating by parts.
Activity 5.11.
Evaluate each of the following indefinite integrals, using the provided hints.
(a) Evaluate
arctan(x) dx by using Integration by Parts with the substitution
u = arctan(x) and dv = 1 dx.
(b) Evaluate
ln(z) dz. Consider a similar substitution to the one in (a).
(c) Use the substitution z = t 2 to transform the integral
t 3 sin(t 2 ) dt to a new
integral in the variable z, and evaluate that new integral by parts.
(d) Evaluate
s 5 e s 3 ds using an approach similar to that described in (c).
(e) Evaluate
e 2t cos(e t ) dt. You will find it helpful to note that e 2t = e t · e t .
⊳
5.4. INTEGRATION BY PARTS
see the possibility for the substitution u = arctan(x) and dv = 1 dx. We explore this
substitution further in Activity 5.11.
In a related problem, if we consider
t 3 sin(t 2 ) dt, two key observations can be made
about the algebraic structure of the integrand: there is a composite function present
in sin(t 2 ), and there is not an obvious function-derivative pair, as we have t 3 present
(rather than simply t) multiplying sin(t 2 ). This problem exemplifies the situation where
we sometimes use both u-substitution and Integration by Parts in a single problem. If we
write t 3 = t · t 2 and consider the indefinite integral
t · t
2 · sin(t
2 ) dt,
we can use a mix of the two techniques we have recently learned. First, let z = t 2 so
that dz = 2t dt, and thus t dt =
1
2 dz. (We are using the variable z to perform a “zsubstitution” since u will be used subsequently in executing Integration by Parts.) Under
this z-substitution, we now have
t · t
2 · sin(t
2 ) dt =
z · sin(z) ·
1
2
dz.
The remaining integral is a standard one that can be evaluated by parts. This, too, is
explored further in Activity 5.11.
The problems briefly introduced here exemplify that we sometimes must think creatively
in choosing the variables for substitution in Integration by Parts, as well as that it is entirely
possible that we will need to use the technique of substitution for an additional change of
variables within the process of integrating by parts.
Activity 5.11.
Evaluate each of the following indefinite integrals, using the provided hints.
(a) Evaluate
arctan(x) dx by using Integration by Parts with the substitution
u = arctan(x) and dv = 1 dx.
(b) Evaluate
ln(z) dz. Consider a similar substitution to the one in (a).
(c) Use the substitution z = t 2 to transform the integral
t 3 sin(t 2 ) dt to a new
integral in the variable z, and evaluate that new integral by parts.
(d) Evaluate
s 5 e s 3 ds using an approach similar to that described in (c).
(e) Evaluate
e 2t cos(e t ) dt. You will find it helpful to note that e 2t = e t · e t .
⊳