40
3.5
Chapter 3. A rapid review of integral calculus
Integration by parts
Consider an indefinite integral
Z
u(x)v 0 (x)dx.
I(x) =
Using the substitution rule, we can re-write this as
Z
I=
But consider
d
dx
u dv.
dv
du
uv) = u
+v .
dx
dx
Integrate both sides (omitting the constant of integration):
Z
uv =
or
du
u dx +
dx
Z
Z
Z
uv =
v
udv +
du
dx,
dx
vdu.
Hence,
Z
I=
Z
udv = uv −
vdu.
(3.7)
In some situations, the integral on the RHS is doable while the one on the LHS is not. Then, we
do the relatively easy integral instead of the more difficult one.
Example: Use integration by parts to find I =
R
xe−x dx.
Solution: Identify u = x and dv = e−x dx. Hence,
v = −e−x .
du = dx,
We have
Z
Z
udv = uv −
vdu = −xe
−x
Z
+
e−x dx = −xe−x − e−x + Const.
Hence,
I(x) = −xe−x − e−x + Const.
Check:
dI
= xe−x = Integrand.
dx
3.5. Integration by parts
41
Example: Use integration by parts to find I =
R
x cos(x) dx.
Solution: Identify u = x and dv = cos(x) dx. Hence,
Z
du = dx,
v=
cos(x) dx = sin(x),
and
Z
I = uv −
vdu,
Z
x sin(x) − sin(x) dx,
x sin(x) + cos(x) + Const.
Example: Use integration by parts to find I =
R
log(x) dx.
Solution: Identify u = log(x) and dv = dx. Hence,
Z
1
du = dx,
x
v=
dx = x
and
Z
I = uv −
vdu,
Z
= x log(x) − x(1/x)dx,
Z
= x log(x) − dx,
= x log(x) − x + Const.
hence
Z
log(x)dx = x log x − x + Const.
(3.8)
Sometimes, integration by parts needs to be done repeatedly before the final answer is obtained, as
in the following example:
Example: Evaluate I =
R π/2
0
x2 cos(x) dx.
42
Chapter 3. A rapid review of integral calculus
Solution: Identify u = x2 and dv = cos(x) dx. Hence,
Z
du = 2x dx,
v=
cos(x) dx = sin(x)
Call the pertinent indefinite integral I(x) –
Z
I(x) =
x2 cos(x) dx.
We have
Z
I = uv −
vdu,
Z
2
= x sin(x) − 2
x sin(x) dx .
(3.9)
We have to do IBP again, on the term inside the square brackets. So, let u = x and dv = sin(x) dx.
Hence, Hence,
Z
du = dx,
sin(x) dx = − cos(x).
v=
Hence,
Z
[· · · ] = uv −
vdu,
Z
= −x cos(x) + cos(x) dx
Substitute back into Equation (3.9):
2
Z
I(x) = x sin(x) − 2
x sin(x) dx ,
Z
2
= x sin(x) − 2 −x cos(x) + cos(x) dx ,
Z
2
= x sin(x) + 2x cos(x) − 2 cos(x) dx,
= x2 sin(x) + 2x cos(x) − 2 sin(x) + Const.
Evaluate at the endpoints to obtain the definite integral:
π/2 h
I = x2 sin(x) + 2x cos(x) − 2 sin(x) 0 =
π 2
4
i
− 0 + (0 − 0) − 2 (1 − 0) = 14 π 2 − 2.