Integration by Parts - Bruce E. Shapiro

Topic 6
Integration by Parts
Integration by parts turns the product rule into a rule for integration. Recall
that from the product rule we had:
(uv) ′ = uv ′ + u ′ v (6.1)
Rearranging and writing this in the Leibniz notation
u dv
dx = d(uv)
dx
− v du
dx
Multiply both sides by the differential dx gives:
Cancelling the dx everywhere,
(6.2)
u dv d(uv)
dx =
dx dx
dx − v du dx (6.3)
dx
udv = d(uv) − vdu (6.4)
Integrate both sides of the equation:
∫ ∫ ∫
udv = d(uv) −
vdu (6.5)
The first term is the integral of an exact differential, so we arrive at the
following result.
Integration by Parts Formula
∫
∫
udv = uv −
vdu (6.6)
« 2012. Last revised: February 26, 2013 Page 15

Math 150A TOPIC 6. INTEGRATION BY PARTS
∫
Example 6.1
x sin x dx
Let
u = x =⇒ dv = dx (6.7a)
∫
dv = sin x =⇒ v = sin xdx = − cos x (6.7b)
Then
∫
∫
x sin x dx = uv − v du
∫
= (x)(− cos x) −
(6.8a)
(− cos x)dx (6.8b)
= −x cos x + sin x + C (6.8c)
∫
An analogous formula holds for x cos x:
∫
x cos x dx = cos x + x sin x + C (6.9)
∫
Example 6.2 xe x dx
Let
Then
u = x =⇒ du = dx (6.10a)
dv = e x dx =⇒ v = e x (6.10b)
∫
∫
xe x dx = (x)(e x ) − (e x )dx (6.11a)
∫
Example 6.3
Let
ln x dx
= xe x − e x + C (6.11b)
Then
∫
u = ln x
=⇒ du = dx x
(6.12a)
dv = dx =⇒ v = x (6.12b)
∫
ln x dx = x ln x −
( ) dx
x = x ln x − x + C (6.13)
x
Page 16 « 2012. Last revised: February 26, 2013

TOPIC 6. INTEGRATION BY PARTS Math 150A
When there are limits of integration we apply them at each step of the
formula.
Example 6.4
∫ 2
1
x ln x dx Let
u = ln x
=⇒ du = dx x
(6.14a)
dv = xdx =⇒ v = 1 2 x2 (6.14b)
Then
∫ 2
1
( )∣ 1 ∣∣∣
2
x ln x dx = (ln x)
x x2
= 2 ln 2 − 1 2
∫ 2
1
= ln 2 2 − 1 4 x2 ∣ ∣∣∣
2
1
1
−
xdx
∫ 2
1
( 1
2 x2 ) ( dx
x
= ln 4 − 1 4 × 4 + 1 4 × 1 = ln 4 − 3 4
)
(6.15a)
(6.15b)
(6.15c)
(6.15d)
Sometimes integration by parts will return the original function to be integration,
but reversed in sign, or multiplied by a constant. When this
happens, one can write an algebraic equation for the result, as illustrated
in the following example.
∫
Example 6.5 e x sin x dx Let
u = e x =⇒ du = e x dx (6.16a)
dv = sin x dx =⇒ v = − cos x (6.16b)
Then
∫
uv− ∫ vdu
{ }} ∫ {
e x sin x dx = −e x cos x + e x cos x dx (6.17)
The second integral on the right doesn’t look any better than what we
started with; nevertheless, we can continue, integrating by parts a second
time, with:
u = e x =⇒ du = e x dx (6.18a)
dv = cos x dx =⇒ v = sin x (6.18b)
« 2012. Last revised: February 26, 2013 Page 17

Math 150A TOPIC 6. INTEGRATION BY PARTS
Then
∫
uv− ∫ vdu
{(
∫}} ){
e x sin x dx = −e x cos x + e x sin x − e x sin x dx
(6.19)
∫
The term with e x sin x dx appears on both sides of the equation, but
with different signs. We can add this to both sides of the equation and
divide by two to get the following:
∫
e x sin x dx = 1 2 ex (sin x − cos x) (6.20)
A common trick is to combine the method of “u-substitution” with the
method of integration by parts, as in the following example.
∫
Example 6.6 cos √ x dx We first make the substitution z = √ x, so that
dz =
dx
2 √ x . Then since dx = 2√ xdz = 2zdz, we have
∫
cos √ ∫
x dx = 2 z cos z dz
Example 6.7
(6.21a)
= 2 (cos z + z sin z) + C by equation 6.9 (6.21b)
= 2 cos √ x + 2 √ x sin √ x + C (6.21c)
∫ √ π
0
∫ √ π
0
x 3 sin(x 2 )dx First substitute z = x 2 , so that
x 3 sin(x 2 )dx =
= 1 2
∫ π
0
∫ π
0
z 3/2 sin z
z sin z dz
dz
2 √ z
The last integral is exactly like the integral in example 6.1, so that
(6.22a)
(6.22b)
∫ √ π
x 3 sin(x 2 )dx = 1 ∣ ∣∣∣
π
2 (−z cos z + sin z) = π 2
0
0
(6.22c)
Page 18 « 2012. Last revised: February 26, 2013

TOPIC 6. INTEGRATION BY PARTS Math 150A
Some of the formulas that we have derived using integration by parts in this
section arise so frequently that it is helpful to jot them down in a separate
table. (Some of these formulas are derived in the exercises.)
∫
1. x sin x dx = sin x − x cos x
∫
2. x cos x dx = cos x + x sin x
∫
3. xe x dx = e x (x − 1)
∫
4. x 2 e x dx = e x (2 − 2x + x 2 )
∫
5. ln x dx = x ln x − x
6.
7.
8.
9.
∫
∫
∫
∫
x ln x dx = x2
2
(
ln x − 1 )
2
sin −1 x dx = √ 1 − x 2 + x sin −1 x
cos −1 x dx = − √ 1 − x 2 + x cos −1 x
tan −1 x dx = x tan −1 x − ln √ 1 + x 2
« 2012. Last revised: February 26, 2013 Page 19

Math 150A TOPIC 6. INTEGRATION BY PARTS
Exercises.
Use integration by parts to solve the following integrals.
∫
6.1. tan −1 x dx
∫
6.2. x tan −1 x dx
∫
6.3. sin −1 (3x) dx
∫
6.4. t 2 e t dt
6.5.
6.6.
6.7.
6.8.
6.9.
6.10.
6.11.
∫ ln x
√ x
dx
∫
∫
∫
∫
∫
∫
e x cos x dx
x2 x dx
x ln(2x) dx
x 2 ln(5x) dx
x 3 (x 2 + 7) 3/2 dx
x 5
√
x3 + 5 dx
Make a substitution then use integration by parts to solve the following
integrals.
∫
6.12. (ln(3x)) 2 dx)
∫
6.13. ln √ x dx
∫
6.14. cos ln x dx
∫
6.15. e x sin −1 e x dx
Page 20 « 2012. Last revised: February 26, 2013