Integral Calculus 201-NYB-05
Vincent Carrier
Integration by Parts
Let u and v be functions of a variable x. By the Product Rule,
(uv)0 = u0 v + uv 0 = uv 0 + u0 v.
Using differentials, this can be written as
d(uv) = u dv + v du.
Integrating on both sides, we get
Z
Z
Z
d(uv) =
u dv + v du
Z
uv =
Z
u dv +
v du.
Therefore,
Z
Z
u dv = uv −
v du.
Integration by parts consists in using the above formula to solve an integral.
The integral on the left is the original integral. The idea is to decompose the integrand
in a product u dv, and then apply the formula. If u and dv are carefully chosen, then the
integral
Z
v du
on the right side should be easier to solve than the original integral
Z
u dv.
If that is not the case, then u and dv should be chosen differently.
Integration by parts is mostly used when the integrand is a product, as the following
examples show.
Examples:
Z
a)
x sin x dx
Z
Z
x sin x dx = −x cos x +
u = x
dv = sin x dx
du = dx
v = − cos x
cos x dx
= −x cos x + sin x + C
Z
b)
ln x
dx
x2
Z
Z
c)
Z
ln x
ln x
dx = −
+
2
x
x
Z
u = ln x
1
dx
x2
= −
ln x 1
− +C
x
x
= −
ln x + 1
+C
x
du =
1
dx
x
1
dx
x2
1
v = −
x
dv =
x2 e2x dx
x2 e2x
x2 e2x dx =
−
2
Z
u = x2
xe2x dx
du = 2x dx
2x
Z
x2 e2x
xe
1
2x
=
−
−
e dx
2
2
2
2x
x2 e2x
xe
e2x
=
−
−
+C
2
2
4
=
dv = e2x dx
(2x2 − 2x + 1)e2x
+C
4
v =
u = x
du = dx
e2x
2
dv = e2x dx
v =
e2x
2
Integration by parts should NOT be used when a basic formula (plus) or a simple substitution would do:
Z
Z
Z
Z
sin x
(ln x)2
2 x3
dx,
dx.
sec x tan x dx,
x e dx,
x
cos5 x
It is also used sometimes when the integrand is a simple function.
Examples:
Z
a)
ln x dx
Z
u = ln x
Z
ln x dx = x ln x −
dx
du =
dv = dx
1
dx
x
v = x
= x ln x − x + C
b) arcsin 3x dx
Z
Z
arcsin 3x dx = x arcsin 3x − 3
√
x
dx
1 − 9x2
u = arcsin 3x
dv = dx
3
du = √
dx
1 − 9x2
1
= x arcsin 3x +
6
Z
u = 1 − 9x2
u−1/2 du
1 u1/2
= x arcsin 3x +
+C
6 1/2
√
= x arcsin 3x +
v = x
1 − 9x2
+C
3
−
1
du = x dx
18
Sometimes, a simple substitution must be used before integration by parts can be applied.
Examples:
Z
a)
√
e
x
dx
Z
√
e
x
Z
dx = 2
t =
tet dt
√
x
2t dt = dx
Z
t
t
= 2 te − e dt
u = t
dv = et dt
du = dt
v = et
= 2 tet − et + C
= 2et (t − 1) + C
= 2e
Z
b)
Z
√
x
√
( x − 1) + C
θ3 cos θ2 dθ
1
θ3 cos θ2 dθ =
2
1
=
2
t = θ2
Z
t cos t dt
1
dt = θ dθ
2
Z
t sin t − sin t dt
=
1
(t sin t + cos t) + C
2
=
1 2
θ sin θ2 + cos θ2 + C
2
u = t
du = dt
dv = cos t dt
v = sin t
The letter t is used for the simple substitution to avoid confusion in the formula for the
integration by parts.
These two integrals could also be solved by first using integration by parts, and then a
simple substitution.
The following integral is an interesting special case.
Z
x
Z
x
e sin x dx = e sin x −
dv = ex dx
u = sin x
ex cos x dx
du = cos x dx
v = ex
Z
x
x
= e sin x − e cos x + e sin x dx
x
u = cos x
dv = ex dx
du = − sin x dx
x
x
Z
= e sin x − e cos x −
Z
2
v = ex
ex sin x dx
ex sin x dx = ex sin x − ex cos x
=
ex (sin x − cos x)
+C
2
As can be seen, the idea consists in:
1) Applying integration by parts twice until the same integral is obtained;
2) Putting both integrals on the same (left) side and isolating.
Integrals of the type
Z
ax
e sin bx dx
Z
and
where a, b ∈ R, are solved using this method.
eax cos bx dx
Definite Integrals
Here are two ways of dealing with a definite integral when using integration by parts.
Consider the definite integral
Z
π/2
x cos x dx.
0
Method 1:
Z
π/2
h
x cos x dx =
x sin x
iπ/2
Z
−
0
0
π/2
sin x dx
0
u = x
du = dx
dv = cos x dx
v = sin x
h
iπ/2
π
sin − 0 sin 0 + cos x
=
2
2
0
π π
=
+ cos − cos 0
2
2
π
π
π−2
−1 =
2
2
=
Method 2:
Z
u = x
Z
x cos x dx = x sin x −
du = dx
= x sin x + cos x + C
Z
π/2
x cos x dx =
h
x sin x + cos x
iπ/2
0
0
=
=
π
2
sin
dv = cos x dx
sin x dx
π
π
+ cos
− (0 sin 0 + cos 0)
2
2
π
π−2
−1 =
2
2
v = sin x