Download Integration By Parts Solutions

Integration By Parts Solutions
Practice Problems: Evaluate the following Integrals
Z
(1)
Z
(2)
Z
ln(x)
dx
x2
(4)
Z
x
e sin(x) dx
Z
(3)
x ln(x + 1) dx
(5)
1
Z
ln(x + 1) dx
(6)
x4 ex dx
x sec2 (x) dx
0
Solutions
(1) Evaluate
Z
ln(x)
dx.
x2
Intuition: Even though the integrand is not in one of the “desired forms” for IBPs, we observe that we
can easily antidifferentiate x12 and ln(x) becomes simpler upon differentiation. With this in mind, we apply
IBPs with u = ln(x) and dv = x12 .
u = ln(x) −→ v = − x1
IBPs gives
Using
du = x1 dx
.
dv = x12 dx
Z
Z
ln(x)
ln(x)
1
dx
=
−
+
dx
x2
x
x2
ln(x) 1
=−
− + C.
x
x
See solution video
Z
(2) Evaluate
ex sin(x) dx.
Intuition: For the integrand ex sin(x) we can choose either function (ex or sin(x)) to play the role of u.
However, make sure that you remain consistent with your choice when you apply IBPs a second time.
u = ex
−→ v = − cos(x)
Using
IBPs gives
du = ex dx
.
dv = sin(x)dx
Z
Z
x
x
e sin(x) dx = −e cos(x) + ex cos(x) dx
Z
= −ex cos(x) + ex sin(x) − ex sin(x) dx
R
where we have applied a second IBPs (u = ex and dv = cos(x)) to ex cos(x) dx. Simplifying algebraically
yields
Z
2 ex sin(x)dx = ex sin(x) − ex cos(x)
so that
ex sin(x) − ex cos(x)
+ C.
2
After the second IBPs we are left with an algebraic equation. An easier, but similar, problem would be to
solve the equation x = a − x for x. Combine like terms and isolate.
Z
ex sin(x) dx =
1
Calculus II Resources
Integration Techniques
See solution video
Z 1
(3) Evaluate
ln(x + 1) dx.
0
Intuition: Similar to the approach used in Example 2 of the IBPS section, we express the integrand as
ln(x + 1) · 1 and let u = ln(x + 1) and dv = 1.
u = ln(x + 1) −→ v = x
IBPs gives
Using
1
dx
.
dv = 1dx
du = x+1
Z
0
1
1 Z 1
x
ln(x + 1) dx = x ln(x + 1) −
dx
0 x+1
0
Z 1
x+1−1
= ln(2) − 0 −
dx
x+1
0
Z 1
1
1−
= ln(2) −
dx
x
+
1
0
1
= ln(2) − x + ln |x + 1|
0
= ln(2) − 1 + ln(2)
= 2 ln(2) − 1.
See solution video
Z
(4) Evaluate
x ln(x + 1) dx.
Observation: Typically when applying IBPs to an integrand of the form xf (x) one would choose u = x
and dv = f (x). Whatever function plays the role of dv MUST be something we can integrate. In this case,
one would need a separate IBPS (See Practice Problem 3) to integrate ln(x + 1). With this in mind, we
choose u = ln(x + 1) and dv = x.
Using
u = ln(x + 1)
1
du = x+1
dx
v = 12 x2
IBPs gives
dv = xdx
−→
.
Z
x ln(x + 1) dx =
1
1 2
x ln(x + 1) −
2
2
Z
x2
dx.
x+1
In order to evaluate the resulting integral we use the substitution w = x + 1, so that dw = dx and x = w − 1.
Z
x ln(x + 1) dx =
=
=
=
=
=
1 2
x ln(x + 1) −
2
1 2
x ln(x + 1) −
2
1 2
x ln(x + 1) −
2
1 2
x ln(x + 1) −
2
1 2
x ln(x + 1) −
2
1 2
x ln(x + 1) −
2
Z
1
x2
dx
2
x+1
Z
1
(w − 1)2
dw
2
w
Z 2
1
w − 2w + 1
dw
2
w
Z
1
1
w − 2 + dw
2
w
1 2
1
w + w − ln |w| + C
4
2
1
1
2
(x + 1) + (x + 1) − ln |x + 1| + C
4
2
2
Calculus II Resources
Integration Techniques
Observation: Alternatively, one could start by first making the substitution w = x + 1, and then apply
IBPs to the resulting integral.
See solution video
Z
(5) Evaluate
x4 ex dx.
Observation: We use the technique of multiple IBPs outlined in Example 4 of the IBPs section.
sign
+
+
+
STOP
u
x4
4x3
12x2
24x
242
0
·
·
·
·
·
dv = ex
ex
ex
ex
ex
ex
We multiply the entries in each row and put the sign term in front of each product to get
Z
x4 ex dx = x4 ex − 4x3 ex + 12x2 ex − 24xex + 24ex + C.
See solution video
Z
(6) Evaluate
x sec2 (x) dx.
Using
u=x
du = dx
−→
.
v = tan(x)
IBPs gives
dv = sec2 (x)dx
Z
x sec2 (x) dx = x tan(x) −
Z
tan(x) dx
= x tan(x) + ln | cos(t)| + C
where we have used Practice Problem 3 from the Substitution Section to evaluate
See solution video
3
R
tan(x) dx.