Chapter 9 Calculus Techniques
for the Elementary Functions
9.2 Integration by parts – A way to integrate
Products
Integration by parts is a rule that transforms the integral of
products of functions into other, hopefully simpler,
integrals.
x
x
e
cos
xdx

e

 dx   cos xdx
We must look at the product rule for derivatives.
Integration by parts – A way to integrate Products
If y  uv where u and v are differentiable
functions of x then:
dy  du  v  u  dv
 dy   vdu   udv
 udv   dy   vdu
 udv  y   vdu
 udv  uv   vdu
Formula for integration by parts.
Integration by parts – A way to integrate Products
Suppose we want to find the volume of the solid
formed by rotating the region under y  cos x
around the y-axis.
dV  2x  ydx
dV  2x cos xdx
V  2  x cos xdx
From aside:



V  2  x sin x  cos x 02 


Aside
Let
 udv  uv   vdu
ux
du  1dx
dv  cos x
v  sin x
 x cos x  x sin x   sin xdx
 x cos x  x sin x  cos x
Integration by parts – A way to integrate Products
Examples:
1.
x
2
sin xdx
x
2.
 udv  uv   vdu
2
sin xdx   x cos x    cos x 2 xdx
x
xe
 dx
2
Let u  x
dv  sin xdx
du  2xdx v   cos x
2
9.3 Rapid Repeated Integration by
Parts
Example:
mult dv
u
 x cos 2 xdx
du
v
U
dV
5
integrate
 udv uv   vdu
Reason for the sign
change.


uv  uv  uv   vdu
uv  uv   vdu
x5
+
5x4
-
20 x 3
+
60 x 2
120 x
120
-
0
+
+
cos 2 x
1
sin 2 x
2
1
 cos 2 x
4
1
 sin 2 x
8
1
cos 2 x
16
1
sin 2 x
32
1
 cos 2 x
64
5
 x cos 2 xdx 
U
x
dV
5
+
5x4
-
20 x 3
+
60 x 2
120 x
120
-
0
1 5
5
10
15
15
15
x sin 2 x  x 4 cos 2 x  x 3 sin 2 x  x 2 cos 2 x  x sin 2 x  cos 2 x  C
2
4
4
4
4
8
+
+
cos 2 x
1
sin 2 x
2
1
 cos 2 x
4
1
 sin 2 x
8
1
cos 2 x
16
1
sin 2 x
32
1
 cos 2 x
64
 Using Trig properties to make original
reappear.
2
sin
 0.4x dx
sin 0.4 x
0.4 cos 0.4 x
+
-
sin 0.4 x
1

cos 0.4 x
0.4
1
2
sin
0
.
4
x
dx


cos
0
.
4
x
sin
0
.
4
x

cos
0.4 x


0.4
1
2
2
sin
0
.
4
x
dx


cos
0
.
4
x
sin
0
.
4
x

1

sin
0.4 x


0.4
1
2
2
sin
0
.
4
x
dx


cos
0
.
4
x
sin
0
.
4
x

1
dx

sin
0.4 xdx



0.4
1
2  sin 2 0.4x dx  
cos 0.4 x sin 0.4 x  x
10.4
2
 sin 0.4x dx   0.8 cos 0.4 x sin 0.4 x  0.5x  C
2


 Reassociation between Steps
 x e dx
5 x
3
x
3
2 x3
xe
1 x3
e
3
+
3x 2
3
0
+
1 2 x3
xe
3
1 x3
e
9
1 3 x3 1 x3
 x e dx  3 x e  3 e  C
5 x3
Integrating with
substitution
method
reassociate
n
cos
 xdx 
1
n 1
n2
cos n 1 x sin x 
cos
xdx

n
n
9.4 Reduction Formulas
 Objective is to find a generic formula for
n
cos
 xdx
 cos
6
xdx
cos5 x
cos x
 5 cos 4 x sin x
sin x
 cos 5 x sin x  5 cos 4 x sin 2 xdx


 cos 5 x sin x  5 cos 4 x 1  cos 2 x dx
 cos 5 x sin x  5 cos 4 xdx  5 cos 6 xdx
1
5
 cos 5 x sin x   cos 4 xdx
6
6
9 – 6 Integration by Trig Substitution
Reminder:
1.

49  x 2 dx1
cos 2 x  1  cos 2 x 
2
1
sin 2 x  1  cos 2 x 
2
x
49  x 2
sin  
cos  
7
7
x  7 sin  sin 2 x   2 sin x cos x 7 cos   49  x 2
dx  7 cos d
2
Since this
cosreminds
2 x   cosus
xof Pythagoras,
sin 2 x
  7label
cos da triangle and attempt to
 7 coslet’s
change into integration of trig.
1
 d  49 1  cos 2 d
 49 cos
Key to success is2deciding whether
49
the variable
x
or 2
the
1 legcos
 is the
 d
hypotenuse. 2
If we choose49
x as a leg1we can choose



sin
2


C
which leg.Vertical
or
horizontal
will


2
2

affect which 49
trigfunction
49 we will use
  2 sin  cos    C
later on. 
2
4
49 1 x 49 x 49  x 2
 sin
  
C
2
7 2 7
7
2
7

49  x 2
x
9 – 6 Integration by Trig Substitution
2
2.
dx
0 4  x 2
When changing from
dx to d must also
change the bounds.
9 – 7 Integration of Rational Functions by Partial
Fractions
Polynomial consta nt consta nt
consta nt


 ... 
Polynomial
linear
linear
linear
A
B
10 x  15
10 x  15



2
x  1 x  10
x  1x  10
x  11x  10
xx11010x10x1515 AxAx110Bx B1x  10
x 10
 10
xx11xx10
  x 1x 1x  10
We are wanting
to ‘undo’ the
common
denominator
10 x 15
BA
x x1 10
Let x=1 so that we get

10x 15
A 
 B rid of B and just have
x  10x  1 x 10


x  19
A left9
10 x  15
 25
115
101  1510
10  15 
2
 A  x  1 B  x  10 
x  11x1 1010
 10  1
SHORTCUT!!
10 x  15
A
B


x  1x  10 x  1 x  10
Simply place finger over (x-1) when letting x=1 and
solve to get A. Then place finger over (x-10) when
solving for B.
Integration of Rational Functions by Partial Fractions
Examples:
1.
2x  6
 x3  x 2  6 x dx
px 
This is an example of a rational function
in proper form. That is px 
qx 
lower degree than
Improper form
2.
3.
q x  .
d egree of px  d egree of qx
x3  6 x 2  11x  1
dx

x 1
x4  1
 xx  1x  1 dx
With improper form we
need to perform long
division first. Partial
fractions will work only
with proper form.
Integration of Rational Functions by Partial Fractions
3x  9
3
6
dx


2
2
 xencounter
isx that
One problem that we may
1 when
 1
x  1 we factor the
denominator, we find some factors repeated, that is occurring with
6
multiplicities greater than 1.
 3 ln x  1 
C
x 1
Repeated linear factors
4.
3x  9
 x  12 dx
Trick to partial fractions:
3x  9
A
B


x  12 x  1 x  12
x  12 3x  9  x  12 A  x  12 B
x  1 x  12
x  12
3x  9  x  1A  B
3x  9  x  1A  6
3x  3  Ax  1
3x  1  Ax  1
6B
3 A
Integration of Rational Functions by Partial Fractions
Unfactorable Quadratics:
5.
x2  4
 x
2

 2 x  1
x2  4
Ax  B
C


x 2  2 x  1 x 2  2 x  1

Trick:
dx

x 2  4   Ax  B x  1  C x 2  2


x 2  4  Ax 2  Ax  Bx  B  Cx 2  2C
x 2  4   A  C x 2   A  B x   B  2C 
1 AC
0  A B
 4   B  2C
x2  4
2x  2 1
 x 2  2 x 1 dx   x 2  2  x 1
2x
2
1
 2
dx   2
dx  
dx
x 2
x 2
x 1


9 – 8 Integrals of the Inverse Trig Functions
“Like climbing mount Everest, they are interesting
more from the standpoint that they can be done,
rather than because they are of great practical
use.”
In 4 – 5 we found derivatives of inverse trig by
implicit differentiation.
d
1
1
sin x 
dx
1 x2
d
1
tan 1 x 
dx
1 x2
d
1
sec 1 x 
dx
x x2 1
d
1
1
cos x  
dx
1 x2
d
1
cot 1 x  
dx
1 x2
d
1
csc 1 x  
dx
x x2 1
9 – 8 Integrals of the Inverse Trig Functions
Integrate using parts.
cot
x

1
1

1 x2
x
 cot x  x cot x   1  x 2 dx
1
1
1
2
cot
x

x
cot
x

ln
1

x
C

2
1
cot 1 x
1
HW 1 – 9 odd
+
-
1
x
Aside: our calc and the
back of the book use
the notion sgnx
x
sgn x 
x
Treat this as a constant
even though we do not
know the sign.
9 – 10 Improper Integrals
Improper Integrals:
•Upper or lower limit of integration is infinite.
•Integrand is discontinuous for at least one value of x at or between the
limits of integration.
An improper integral converges to a certain number if each applicable limit
shown below is finite. Otherwise, the integral diverges.

b
 f ( x)dx  lim  f ( x)dx
a
b 
a
b
b
 f ( x)dx  lim  f ( x)dx

a  
b
a
b
k
 f ( x)dx  lim  f ( x)dx  lim  f ( x)dx,
a
k c 
k
k c 
a
f Is discontinuous at x=c in
a, b
Improper Integrals
** If the integrand seems to approach zero as x gets very large or small,
1
1
then
the
converge. Conversely, if the integrand grows
1 integral might
0.7
dx  limas xx grows
dx without bound, then the integral definitely
without
0.7 bound
a

0
x
0
a
diverges.

Examples:

1
1
dx,
1. For the improper integral  x 0.7
0
a. Graph the integrand, and tell whether or not the integral might
converge.
Graph looks to be approaching 0
and therefore might converge.
b. If the integral might converge,
find out whether or not it really does,
and if so, to what limit it converges.
Improper Integrals
3
2. For the improper integral
1
3 x 2 dx,
a. Graph the integrand, and tell whether or not the integral might
converge.
b. If the integral might converge, find out whether or not it really does,
and if so, to what limit it converges.

3. For the improper integral
dx
3 x 4 ,
a. Graph the integrand, and tell whether or not the integral might
converge.
b. If the integral might converge, find out whether or not it really does,
and if so, to what limit it converges.
9.11 Miscellaneous Integrals and
Derivatives
Differentiation
 Sum u  v '  u'v'
 Product
u  v'  u' v  uv'
 Quotient  u 
u ' v  uv'
 ' 
v





v2
Composite  f u '  f ' u  u'
Implicit f  y   g x  f '  y  y'  g ' x
n
Power Function x '  nx n 1
x
Exponential Function n x '  n ln n
Logarithmic function log b x '  1
x ln b
 
 
Differentiation Con’t
 Logarithmic y  f x
ln y  ln f x

1
y '  ln f x '
y
 y'  yln f x'
 Trigonometric Function sin' x  cos x
cos' x   sin x
tan' x  sec 2 x
 Inverse Trig Function – differentiate
Implicitly
Integration
 Sum
 u  v dx
 Product
 udv
  udx   vdx
 uv   vdu
 Reciprocal Function
 Power Function
1
u
 du
 ln u  C
n
u
 du  1 u n1  C
 Power of a Function
n 1

f n  x dx
n  1
reduction formula
Integration Continued
 Square root of quadratic: Trig Sub
 Rational Algebraic Expression:
Convert to sum by long division,
partial fractions
 Inverse Function: Integrate by parts