Summary of Patterns in Integration by Parts
The General Statement: Integration by parts comes from a “Backwards Product Rule”:
Z
Z
u dv = uv − v du
Z
Pattern 1:
xn f (x) dx
In this case, make a table, using u = xn , dv = f (x) dx. Differentiating the u column will give you zero at
some point. ZWorks when f (x) is easily antidifferentiated (like ekx , sin(kx), cos(kx)).
x3 e2x dx
Example:
Sign u
+
x3
−
3x2
+
6x
−
6
+
0
Z
Pattern 2:
Z
ln(x) dx,
dv
e2x
3
Z
x
3x2
6x
6
(1/2)e2x
3 2x
⇒ x e dx =
−
+
−
e2x + C
(1/4)e2x
2
4
8
16
(1/8)e2x
(1/16)e2x
sin−1 (x) dx,
Z
tan−1 (x) dx
−1
The idea for this one comes from the fact that the derivatives of ln(x), sin−1
R (x), tan (x) are all
R algebraic
expressions of
x.
The
general
technique
is
to
let
u
=
f
(x),
and
dv
=
dx,
so
that
f
(x)
dx
=
xf
(x)−
xf 0 (x) dx
Z
Example:
ln(x) dx
Sign
u
+
ln(x)
−
(1/x)
Z
Pattern 3:
ekx sin(nx) dx,
Z
Z
Z
dv
1
1 ⇒ ln(x) dx = x ln(x) − x dx = x ln(x) − x + C
x
x
ekx cos(mx) dx
The idea for this one comes from the fact that the derivatives of ekx give back a constant times ekx ,
coupled with the second derivative/antiderivative of sin(nx) or cos(mx) giving back a constant times sin(nx)
or cos(mx) (respectively). So, use integration by parts twice (using a table) to get the same integral on both
sides of the Zequation, then solve for the integral.
Example:
e2x cos(3x) dx
Sign
u
dv
Z
Z
1 2x
2 2x
4
+
e2x
cos(3x)
2x
⇒
e
cos(3x)
dx
=
e
sin(3x)
+
e
cos(3x)
−
e2x cos(3x) dx
−
2e2x (1/3) sin(3x)
3
9
9
+
4e2x −(1/9) cos(3x)
Z
4
Now add
e2x cos(3x) dx to both sides:
9
Z
13
1
2
e2x cos(3x) dx = e2x sin(3x) + e2x cos(3x)
9
3
9
To get a final answer:
Z
e2x cos(3x) dx =
3 2x
2
e sin(3x) + e2x cos(3x) + C
13
13
1