1.5. INTEGRATION BY PARTS
21
1.5. Integration by Parts
The method of integration by parts is based on the product rule for
differentiation:
[f (x)g(x)]0 = f 0 (x)g(x) + f (x)g 0 (x) ,
which we can write like this:
f (x)g 0 (x) = [f (x)g(x)]0 − f 0 (x)g(x) .
Integrating we get:
Z
Z
Z
0
0
f (x) g (x) dx = [f (x)g(x)] dx − g(x)f 0 (x) dx ,
i.e.:
Z
Z
0
f (x)g (x) dx = f (x)g(x) −
g(x)f 0 (x) dx .
Writing u = f (x), v = g(x), we have du = f 0 (x) dx, dv = g 0 (x) dx,
hence:
Z
Z
u dv = uv − v du .
Example: Integrate
R
xex dx by parts.
Answer : In integration by parts the key thing is to choose u and
dv correctly. In this case the “right” choice is u = x, dv = ex dx, so
du = dx, v = ex . We see that the choice is right because the new
integral that we obtain after applying the formula of integration by
parts is simpler than the original one:
Z
Z
x
x
x e dx = x e −
ex dx = xex − ex + C .
|{z} | {z } |{z} |{z}
|{z} |{z}
u
dv
u
v
v
du
Usually it is a good idea to check the answer by differentiating it:
(xex − ex + C)0 = ex + xex − ex = xex .
A couple of additional typical examples:
Z
Example: x sin x dx = · · ·
1.5. INTEGRATION BY PARTS
22
u = x, dv = sin x dx, so du = dx, v = − cos x:
Z
Z
··· =
x sin x dx = x (− cos x) − (− cos x) dx
|{z} | {z } |{z} | {z }
| {z } |{z}
u
u
dv
v
v
du
= −x cos x + sin x + C .
Z
Example:
ln x dx = · · ·
u = ln x, dv = dx, so du = x1 dx, v = x:
Z
Z
1
··· =
ln x dx = ln x x −
dx
x
x
|{z} |{z} |{z} |{z}
|{z} |{z}
u
u
v
v
dv
du
Z
= x ln x − dx
= x ln x − x + C .
Sometimes we need to use the formula more than once.
Z
Example: x2 ex dx = . . .
u = x2 , dv = ex dx, so du = 2x dx, v = ex :
Z
Z
2
x
2 x
··· =
x e dx = x e − ex 2x dx = . . .
|{z} | {z }
u
dv
u = 2x, dv = ex dx, so du = 2dx, v = ex :
Z
Z
2 x
x
2 x
x
2x e dx = x e − 2xe + 2ex dx
··· = x e −
|{z} | {z }
u
dv
= x2 ex − 2xex + 2ex + C .
In the following example the formula of integration by parts does
not yield a final answer, but an equation verified by the integral from
which its value can be derived.
Z
Example: sin x ex dx = . . .
1.5. INTEGRATION BY PARTS
23
u = sin x, dv = ex dx, so du = cos x dx, v = ex :
Z
Z
x
x
· · · = sin x e dx = sin x · e − ex cos x dx = . . .
|{z} | {z }
u
dv
u = cos x, dv = ex dx, so du = − sin x dx, v = ex :
Z
x
. . . = sin x · e − cos x ex dx
| {z } | {z }
u
dv
Z
x
x
= sin x · e − cos x · e − ex sin x dx
Hence the integral I =
R
sin x ex dx verifies
I = sin x · ex − cos x · ex − I ,
i.e.,
2I = sin x · ex − cos x · ex ,
hence
I = 12 ex (sin x − cos x) + C .
1.5.1. Integration by parts for Definite Integrals. Combining the formula of integration by parts with the Evaluation Theorem
we get:
Z b
Z b
b
0
f (x)g (x) dx = [f (x)g(x)]a −
g(x)f 0 (x) dx .
a
a
Z
1
Example:
0
tan−1 x dx = · · ·
1
u = tan−1 x, dv = dx, so du =
dx, v = x:
1 + x2
Z 1
Z 1
1
−1
−1
1
··· =
tan x dx = [ tan x · x ]0 −
x
dx
1 + x2
0 | {z } |{z}
0 |{z} |
| {z } |{z}
{z }
u
dv
u
v
v
£
¤
= tan−1 1 · 1 − tan−1 0 · 0 −
Z 1
π
x
= −
dx
2
4
0 1+x
Z
0
du
1
x
dx
1 + x2
1.5. INTEGRATION BY PARTS
24
The last integral can be computed with the substitution t = 1 + x2 ,
dt = 2x dx:
Z
1
0
1
x
dx
=
1 + x2
2
Z
2
1
ln 2
1
1
dt = [ln t]21 =
.
t
2
2
Hence the original integral is:
Z
1
tan−1 x dx =
0
π ln 2
−
.
4
2
1.5.2. Reduction Formulas. Assume that we want to find the
following integral for a given value of n > 0:
Z
xn ex dx .
Using integration by parts with u = xn and dv = ex dx, so v = ex and
du = nxn−1 dx, we get:
Z
n x
n x
x e dx = x e − n
Z
xn−1 ex dx .
On the right hand side we get an integral similar to the original one
but with x raised to n−1 instead of n. This kind of expression is called
a reduction formula. Using this same formula several
taking
R xtimes, and
x
into account that for n = 0 the integral becomes e dx = e + C, we
can evaluate the original integral for any n. For instance:
Z
3 x
3 x
Z
x2 ex dx
Z
3 x
2 x
= x e − 3(x e − 2 xex dx)
Z
3 x
2 x
x
= x e − 3(x e − 2(xe − ex dx))
x e dx = x e − 3
= x3 ex − 3(x2 ex − 2(xex − ex )) + C
= x3 ex − 3x2 ex + 6xex − 6ex + C .
1.5. INTEGRATION BY PARTS
25
Another example:
Z
Z
n
sin x dx =
sinn−1 x sin x dx
| {z } | {z }
u
n−1
= − sin
dv
Z
x cos x + (n − 1)
= − sinn−1 x cos x + (n − 1)
− (n − 1)
Z
Z
cos2 x sinn−2 dx
| {z }
1−sin2 x
sinn−2 dx
sinn x dx
Adding the last term to both sides and dividing by n we get the following reduction formula:
Z
Z
sinn−1 x cos x n − 1
n
sin x dx = −
+
sinn−2 x dx .
n
n