Letting dv = dx Using Integration by parts twice Recurring Integrals
Letting dv = dx
Example
R2
−2
ln(x + 3)dx.
Z
Z
u dv = uv −
Annette Pilkington
v du.
Integration by Parts
Letting dv = dx Using Integration by parts twice Recurring Integrals
Letting dv = dx
Example
R2
−2
ln(x + 3)dx.
Z
Z
u dv = uv −
I
v du.
Let u = ln(x + 3), dv = dx
du =
1
dx
x +3
Annette Pilkington
and
v = x.
Integration by Parts
Letting dv = dx Using Integration by parts twice Recurring Integrals
Letting dv = dx
Example
R2
−2
ln(x + 3)dx.
Z
Z
u dv = uv −
I
v du.
Let u = ln(x + 3), dv = dx
du =
I
Z
2
−2
1
dx
x +3
and
v = x.
Z
˛2
˛
ln(x + 3)dx = x ln(x + 3)˛ −
−2
Annette Pilkington
Integration by Parts
2
−2
x
dx
x +3
Letting dv = dx Using Integration by parts twice Recurring Integrals
Letting dv = dx
Example
R2
−2
ln(x + 3)dx.
Z
Z
u dv = uv −
I
Let u = ln(x + 3), dv = dx
du =
I
Z
2
−2
I
v du.
1
dx
x +3
and
v = x.
Z
˛2
˛
ln(x + 3)dx = x ln(x + 3)˛ −
−2
2
−2
x
dx
x +3
R2 x
To calculate −2 x+3
dx, we use substitution with w = x + 3. Then
x = w − 3 and w (−2) = 1, w (2) = 5. We get
Z 2
Z 5
x
w −3
dx =
dw
w
−2 x + 3
1
Z 5
Z 5
˛5
3
˛
=
1dw −
dw = 4 − 3 ln |w |˛ = 4 − 3(ln(5)).
1
1
1 w
Annette Pilkington
Integration by Parts
Letting dv = dx Using Integration by parts twice Recurring Integrals
Letting dv = dx
Example
R2
−2
ln(x + 3)dx.
Z
Z
u dv = uv −
I
Z
2
−2
v du.
Z
˛2
˛
ln(x + 3)dx = x ln(x + 3)˛ −
−2
Annette Pilkington
Integration by Parts
2
−2
x
dx
x +3
Letting dv = dx Using Integration by parts twice Recurring Integrals
Letting dv = dx
Example
R2
−2
ln(x + 3)dx.
Z
Z
u dv = uv −
I
Z
2
−2
I
v du.
Z
˛2
˛
ln(x + 3)dx = x ln(x + 3)˛ −
−2
2
−2
x
dx
x +3
˛2
˛
= x ln(x + 3)˛ − (4 − 3 ln(5).)
−2
Annette Pilkington
Integration by Parts
Letting dv = dx Using Integration by parts twice Recurring Integrals
Letting dv = dx
Example
R2
−2
ln(x + 3)dx.
Z
Z
u dv = uv −
I
Z
2
−2
I
v du.
Z
˛2
˛
ln(x + 3)dx = x ln(x + 3)˛ −
−2
2
−2
x
dx
x +3
˛2
˛
= x ln(x + 3)˛ − (4 − 3 ln(5).)
−2
I
= 2 ln(5) + 2 ln(1) − (4 − 3 ln(5).)
= 2 ln(5) − 4 + 3(ln(5)) = 5 ln(5) − 4.
Annette Pilkington
Integration by Parts
Letting dv = dx Using Integration by parts twice Recurring Integrals
Using Integration by parts twice.
Example
R
(ln x)2 dx.
Z
Z
u dv = uv −
Annette Pilkington
v du.
Integration by Parts
Letting dv = dx Using Integration by parts twice Recurring Integrals
Using Integration by parts twice.
Example
R
(ln x)2 dx.
Z
Z
u dv = uv −
I
v du.
Let u = (ln(x))2 , dv = dx
du =
2 ln x
dx
x
Annette Pilkington
and
v = x.
Integration by Parts
Letting dv = dx Using Integration by parts twice Recurring Integrals
Using Integration by parts twice.
Example
R
(ln x)2 dx.
Z
Z
u dv = uv −
I
Let u = (ln(x))2 , dv = dx
du =
I
v du.
Z
2
2
2 ln x
dx
x
Z
(ln x) dx = x(ln x) − 2
Annette Pilkington
and
v = x.
x ln x
dx = x(ln x)2 − 2
x
Integration by Parts
Z
ln x dx
Letting dv = dx Using Integration by parts twice Recurring Integrals
Using Integration by parts twice.
Example
R
(ln x)2 dx.
Z
Z
u dv = uv −
I
Let u = (ln(x))2 , dv = dx
du =
I
Z
2
2
2 ln x
dx
x
Z
(ln x) dx = x(ln x) − 2
I
v du.
and
v = x.
x ln x
dx = x(ln x)2 − 2
x
In a previous example, we saw that
Z
2 ln(x) dx = 2[x ln x − x] + C
Annette Pilkington
Integration by Parts
Z
ln x dx
Letting dv = dx Using Integration by parts twice Recurring Integrals
Using Integration by parts twice.
Example
R
(ln x)2 dx.
Z
Z
u dv = uv −
I
Let u = (ln(x))2 , dv = dx
du =
I
v du.
Z
2
2
2 ln x
dx
x
Z
(ln x) dx = x(ln x) − 2
and
v = x.
x ln x
dx = x(ln x)2 − 2
x
Z
ln x dx
I
In a previous example, we saw that
Z
2 ln(x) dx = 2[x ln x − x] + C
I
Therefore
Z
Z
(ln x)2 dx = x(ln x)2 − ln x dx = x(ln x)2 − 2x ln x − 2x + C
= x(ln x)2 − 2x ln x + 2x + C
Annette Pilkington
Integration by Parts
Letting dv = dx Using Integration by parts twice Recurring Integrals
Example
Example
R
e 2x cos(5x) dx.
Z
Z
u dv = uv −
Annette Pilkington
v du.
Integration by Parts
Letting dv = dx Using Integration by parts twice Recurring Integrals
Example
Example
R
e 2x cos(5x) dx.
Z
Z
u dv = uv −
I
v du.
Let u = e 2x , dv = cos(5x)dx
du = 2e 2x dx
Annette Pilkington
and
v=
sin(5x)
5
Integration by Parts
Letting dv = dx Using Integration by parts twice Recurring Integrals
Example
Example
R
e 2x cos(5x) dx.
Z
Z
u dv = uv −
I
v du.
Let u = e 2x , dv = cos(5x)dx
du = 2e 2x dx
I
and
v=
sin(5x)
5
Z
e 2x cos(5x) dx =
e 2x sin(5x)
−
5
Z
e 2x cos(5x) dx =
e 2x sin(5x)
2
−
5
5
Annette Pilkington
Z
2e 2x
Z
Integration by Parts
sin(5x)
dx
5
e 2x sin(5x) dx
Letting dv = dx Using Integration by parts twice Recurring Integrals
Example
Example
R
e 2x cos(5x) dx.
Z
Z
u dv = uv −
I
Let u = e 2x , dv = cos(5x)dx
du = 2e 2x dx
I
I
v du.
and
v=
sin(5x)
5
Z
e 2x cos(5x) dx =
e 2x sin(5x)
−
5
Z
e 2x cos(5x) dx =
e 2x sin(5x)
2
−
5
5
To calculate
Z
Z
2e 2x
Z
e 2x sin(5x) dx
we use integration by parts again.
Annette Pilkington
Integration by Parts
sin(5x)
dx
5
e 2x sin(5x) dx
Letting dv = dx Using Integration by parts twice Recurring Integrals
Example
Example
R
e 2x cos(5x) dx.
I
Z
e 2x cos(5x) dx =
e 2x sin(5x)
2
−
5
5
Annette Pilkington
Z
Integration by Parts
e 2x sin(5x) dx
Letting dv = dx Using Integration by parts twice Recurring Integrals
Example
Example
R
e 2x cos(5x) dx.
I
Z
e 2x cos(5x) dx =
I
Z
e 2x sin(5x)
2
−
5
5
Z
e 2x sin(5x) dx
e 2x sin(5x) dx
Let u = e 2x and dv = sin(5x)dx.
du = 2e 2x and v = − cos(5x)
.
5
Z
e 2x sin(5x) dx =
−e 2x cos(5x)
2
+
5
5
Annette Pilkington
Z
Integration by Parts
e 2x cos(5x) dx
Letting dv = dx Using Integration by parts twice Recurring Integrals
Example
Example
R
e 2x cos(5x) dx.
I
Z
e 2x cos(5x) dx =
I
Z
e 2x sin(5x)
2
−
5
5
Z
e 2x sin(5x) dx
e 2x sin(5x) dx
Let u = e 2x and dv = sin(5x)dx.
du = 2e 2x and v = − cos(5x)
.
5
Z
e 2x sin(5x) dx =
−e 2x cos(5x)
2
+
5
5
Z
e 2x cos(5x) dx
I
Z
e 2x cos(5x) dx =
e 2x sin(5x)
2 −e 2x cos(5x)
2
− [
+
5
5
5
5
We now have a recurring integral.
Annette Pilkington
Integration by Parts
Z
e 2x cos(5x) dx]
Letting dv = dx Using Integration by parts twice Recurring Integrals
Example
Example
R
e 2x cos(5x) dx.
I
Z
e 2x cos(5x) dx =
e 2x sin(5x) 2 (−e 2x cos(5x))
2
− [
+
5
5
5
5
Annette Pilkington
Integration by Parts
Z
e 2x cos(5x) dx]
Letting dv = dx Using Integration by parts twice Recurring Integrals
Example
Example
R
e 2x cos(5x) dx.
I
Z
I
e 2x cos(5x) dx =
Let I =
R
e 2x sin(5x) 2 (−e 2x cos(5x))
2
− [
+
5
5
5
5
Z
e 2x cos(5x) dx
I =
e 2x sin(5x)
2 2x
4
+
e cos(5x) −
I.
5
25
25
Annette Pilkington
Integration by Parts
e 2x cos(5x) dx]
Letting dv = dx Using Integration by parts twice Recurring Integrals
Example
Example
R
e 2x cos(5x) dx.
I
Z
I
e 2x cos(5x) dx =
Let I =
R
e 2x sin(5x) 2 (−e 2x cos(5x))
2
− [
+
5
5
5
5
e 2x cos(5x) dx
e 2x sin(5x)
2 2x
4
+
e cos(5x) −
I.
5
25
25
I =
I
Let I =
R
Z
e 2x cos(5x) dx
(1 +
e 2x sin(5x)
4
2 2x
)I =
+
e cos(5x)
25
5
25
e 2x sin(5x)
29
2 2x
I =
+
e cos(5x)
25
5
25
Annette Pilkington
Integration by Parts
e 2x cos(5x) dx]
Letting dv = dx Using Integration by parts twice Recurring Integrals
Example
Example
R
e 2x cos(5x) dx.
I
Z
I
e 2x cos(5x) dx =
Let I =
R
e 2x sin(5x) 2 (−e 2x cos(5x))
2
− [
+
5
5
5
5
Let I =
R
e 2x cos(5x) dx]
e 2x cos(5x) dx
e 2x sin(5x)
2 2x
4
+
e cos(5x) −
I.
5
25
25
I =
I
Z
e 2x cos(5x) dx
(1 +
e 2x sin(5x)
4
2 2x
)I =
+
e cos(5x)
25
5
25
e 2x sin(5x)
29
2 2x
I =
+
e cos(5x)
25
5
25
I
Z
I =
e 2x cos(5x) dx =
Annette Pilkington
i
25 h e 2x sin(5x)
2 2x
+
e cos(5x) .
29
5
25
Integration by Parts
Letting dv = dx Using Integration by parts twice Recurring Integrals
Reduction formula for
Example
R
R
sin3 x dx.
sin3 xdx.
Z
Z
u dv = uv −
Annette Pilkington
v du.
Integration by Parts
Letting dv = dx Using Integration by parts twice Recurring Integrals
Reduction formula for
Example
R
R
sin3 x dx.
sin3 xdx.
Z
Z
u dv = uv −
I
v du.
Let u = sin2 x, dv = sin xdx
du = 2 sin x cos xdx
Annette Pilkington
and
v = − cos x.
Integration by Parts
Letting dv = dx Using Integration by parts twice Recurring Integrals
Reduction formula for
Example
R
R
sin3 x dx.
sin3 xdx.
Z
Z
u dv = uv −
I
Let u = sin2 x, dv = sin xdx
du = 2 sin x cos xdx
I
v du.
Z
and
v = − cos x.
Z
sin3 xdx = − cos x sin2 x + 2 (cos x)(sin x cos x) dx
Z
= − cos x sin2 x + 2 cos2 x sin x dx
Annette Pilkington
Integration by Parts
Letting dv = dx Using Integration by parts twice Recurring Integrals
Reduction formula for
Example
R
R
sin3 x dx.
sin3 xdx.
Z
Z
u dv = uv −
I
Let u = sin2 x, dv = sin xdx
du = 2 sin x cos xdx
I
I
v du.
Z
and
v = − cos x.
Z
sin3 xdx = − cos x sin2 x + 2 (cos x)(sin x cos x) dx
Z
= − cos x sin2 x + 2 cos2 x sin x dx
Z
= − cos x sin2 x + 2 (1 − sin2 x) sin x dx
Z
Z
= − cos x sin2 x + 2 sin x dx − 2 sin3 x dx
Annette Pilkington
Integration by Parts
Letting dv = dx Using Integration by parts twice Recurring Integrals
Reduction formula for
Example
R
R
sin3 x dx.
sin3 xdx.
Z
Z
Z
sin3 xdx = − cos x sin2 x + 2 sin x dx − 2 sin3 x dx
Annette Pilkington
Integration by Parts
Letting dv = dx Using Integration by parts twice Recurring Integrals
Reduction formula for
Example
I
R
R
sin3 x dx.
sin3 xdx.
Z
Z
Z
sin3 xdx = − cos x sin2 x + 2 sin x dx − 2 sin3 x dx
Letting I =
R
sin3 x dx, we have
I = − cos x sin2 x + 2
Annette Pilkington
Z
sin x dx − 2I
Integration by Parts
Letting dv = dx Using Integration by parts twice Recurring Integrals
Reduction formula for
Example
I
R
R
sin3 x dx.
sin3 xdx.
Z
Z
Z
sin3 xdx = − cos x sin2 x + 2 sin x dx − 2 sin3 x dx
Letting I =
R
sin3 x dx, we have
I = − cos x sin2 x + 2
I
Z
sin x dx − 2I
Therefore
3I = − cos x sin2 x − 2 cos x + C
Annette Pilkington
Integration by Parts
Letting dv = dx Using Integration by parts twice Recurring Integrals
Reduction formula for
Example
I
R
R
sin3 x dx.
sin3 xdx.
Z
Z
Z
sin3 xdx = − cos x sin2 x + 2 sin x dx − 2 sin3 x dx
Letting I =
R
sin3 x dx, we have
I = − cos x sin2 x + 2
I
Z
sin x dx − 2I
Therefore
3I = − cos x sin2 x − 2 cos x + C
I
and
Z
1
sin3 x dx = I = − [cos x sin2 x + 2 cos x] + C .
3
Annette Pilkington
Integration by Parts