Download July 22 - Penn Math

Calculus Weeks 3&4
Integration by parts
This is another way to try and integrate products.
It is in fact the opposite of the product rule for derivatives:
Product rule for derivatives: d(uv) = u dv + v du .
Integration by parts:
u
dv

uv

v
du


Use this when you have a product under the integral sign, and it
appears that integrating one factor and differentiating the other
will make the resulting integral easier.
Example:
x
e
dx

2 x
-- There's no other rule for this,
so we try parts.
If we differentiate x2 , we'll get 2x which seems simpler.
And integrating ex doesn't change anything.
Let u = x and dv = ex dx. Then du = 2x dx and v = ex.
The integration by parts formula then gives us:
x
e
dx

x
e

2
x
e
dx


2 x
2 x
x
x
e
dx

x
e

2
x
e
dx


2 x
2 x
x
Continuing:
We can use parts again on this integral
(with u = 2x and dv = ex dx) to get:
x
e
dx

x
e

2
x
e

2
e
dx


2 x
2 x
x
x
 x e  2 x e  2e  C
2 x
x
x
You try one...
x
sin
2
x
dx

…contrast this with
x
sin
x
dx

2
XOXXOX
There is a method called the "tic-tac-toe"
method for integration by parts.
Different strategies for choosing u and dv are
called for to integrate x ln x or x tan-1 x
orx something like ex sin x
A problem for you...
p
x cos(x) dx
Find the value of
0
A) p
E) -2
B) 2p
F) 1
C) 2
G) 1/2
D) 0
H) p/2
Last one for today!
1
x ln x dx
Evaluate
A) -1/4
B) -1/2
C) 0
D) 1/4
E) 1/2
0
Trigonomeric substitutions
These are just substitutions like the ordinary
substitution method, but some of them are a little
surprising. There are two kinds:
1 The first, 1
To integrate products of powers of sine and cosine
(like sin 3 (x) cos 6 (x) ) . You need the identities:
sin 2(x) + cos 2(x) = 1 (everybody knows that one!),
1+cos(2x)
and the double angle formulas: cos (x) =
2
and sin2 (x) = 1-cos(2x)
2
2
The trick is as follows...
(a) If both the power of sine and the power of cosine
are even, then use the double angle formulas to
divide both in half. Keep doing this until at least one
of the powers is odd.
(b) Once at least one of the powers is odd, say it's the
power of cosine, then let u = sin( -- ) -- then use the
Pythagorean identity to convert all but one power of
cosine to sines, and the last power of cosine will be
the du in a substitution.
Let’s do one...
sin
x
cos
x
dx

4
4
Both the powers are even, so use the
double angle formula trick:
sin x cos x 
1
16

1
16
4
4
1  cos 2 x  1  cos 2 x 
2
1  2 cos
2
2
2 x  cos 2 x
4

In the last two terms, use the identities again to
get what’s on the next slide.
Trig identity city!
1
16
2


1

cos
4
x


2
4

1  2 cos 2 x  cos 2 x  161 1  1  cos 4 x   



2




2
1
 64 1  2 cos 4 x  cos 4 x




1
3  4 cos 4 x  cos 8 x 
 128
…and (finally!) this is something we can
integrate!
1
sin
x
cos
x
dx


 128 3  4 cos 4 x  cos 8x  dx
4
4
…and at last:
1
sin
x
cos
x
dx

128  3  4 cos 4 x  cos 8 x dx

4
4
1
3x  sin 4 x  18 sin 8x   C
 128
3
1
1
 128
x  128
sin 4 x  1024
sin 8 x  C
Here’s one with an odd power for you to do:
p
2
2
3
sin
x
cos
x
dx


0
A) 2/15
E) 2/3
B) 4/15
F) 4/5
C) 2/5
G) 14/15
D) 8/15
Try this one:
A) 2
B) p
p
4
cos x dx =
0
C) p - 1/2
D) 2 p
E) 3p/8
2
The second,
2
-- and more important, kind of trig substitution happens
when there is a sum or difference of squares in the
integrand. Usually one of the squares is a constant and
the other involves the variable.
If some other substitution doesn't suggest itself first, then
try the Pythagorean trig identity that has the same
pattern of signs as the one in the problem.
Either
cos 2  1  sin 2 , sec 2  1  tan 2 , or tan 2  sec 2  1
Example:

x
If you think about it, the
substitution u = 5 - x2 won’t
work, because of the extra
factor of x in the numerator.
2
5 x
2
dx
But 5 - x2 is vaguely reminiscent of 1 - sin2, so let
x  5 sin 
2
Then
and
2
x  5 sin 
dx  5 cos
5  x  5  5 sin   5 cos 
2
2
2
We make all the substitutions and get:

x
2
5 x
2
dx  
5 sin 
5 cos
2
5 cos d
  5 sin  d
2
This is a trig integral of the other sort . And we can
use the double-angle formula to get:
1  cos 2
5
5
5
sin

d


5
d




2
4 sin 2  C

 2
2
Now we have to get back to x’s
So far,

x2
5 x
2
 52   54 sin 2  C where
x
therefore sin  
5
Also,
and so
x  5 sin 
 x 
  arcsin  
 5
sin 2  2 sin  cos
5
And from the triangle,
5  x2
cos  
5

5  x2
x
The conclusion is:

x2
5 x
2
dx  52   54 sin 2  C
 52   52 sin  cos  C
2 

x 5  x  5  x 
5
 2 arcsin
 2
C

5
5 
 5 
x 1
5
 2 arcsin
 2 x 5  x2  C
5
Quite a bit of work for one integral!
Examples for you…
Here are two for you to work on.
Notice the subtle difference in the integrand that
changes entirely the method used:
x
4

x
dx

2
x

2
4  x dx
2
Partial Fractions
Last but not least is integration by partial fractions. This method
is based on an algebraic trick. It works to integrate a rational
function (quotient of polynomials) when the degree of the
denominator is greater than the degree of the numerator
(otherwise "when in doubt, divide it out") and you can factor
the denominator completely.
In full generality, partial fractions works when the denominator
has quadratic and repeated factors, but we will consider only
the case of distinct linear factors (i.e., the denominator factors
into linear factors and they're all different). It also uses the
(easy) fact that
1
1
 ax  b dx  a ln ax  b   C
Partial Fractions
(continued)
The idea of partial fractions is to take a rational function of the
form
p ( x)
( x  a)( x  b)( x  c)...
(where there can be more or fewer factors in the denominator,
but the degree of p(x) must be less than the number of factors)
and rewrite it as a sum:
A
B
C


 ...
x a x b x c
for some constants A, B, C (etc..).
For instance,
You can rewrite
2x 1
,
2
x  4x  3
first by factoring the denominator:
2x 1
x  1x  3
and then in partial fractions as:
5
 12
 2
x 1 x  3
Two questions...
1. Why do partial fractions?
2. How to do
partial
fractions?
Two reasons for why:
First reason:
It helps us to do integrals.
From the previous example, we see that
we don't know how to integrate the
fraction
2x 1
right away, but ...
x2  4x  3
5
1
2x 1
2
2
dx

dx

 x 2  4 x  3  x  3  x  1 dx
 52 ln( x  3)  12 ln( x  1)  C
Two reasons for why:
Second reason:
Partial fractions helps us to understand the behavior of a rational
function near its "most interesting” points.
2x 1
For the same example, we graph the function 2
in blue,
x  4x  3
 12
and the partial fractions
and
in red:
x 1
x3
5
2
Let’s take a closer look
red and blue curves
Note that one or the other of the red curves mimics
the behavior of the blue one at each singularity.
OK, now for the “how”
First, we give the official version, then a short-cut.
Official version: It is a general fact that the original fraction
will break up into a sum with one term for each factor in the
denominator. So (in the example), write
2x 1
A
B


2
x  4x  3 x  3 x 1
where A and B are to be determined. To determine A and B,
pick two values of x other than x=-3 or x=-1, substitute them into
the equation, and then solve the resulting two equations for the
two unknowns A and B. For instance, we can put x=0, and get
1
3
 13 A  B and for x=1, get
3
8
 14 A  12 B .
Solve to get A = 5/2 and B = -1/2 as we did before.
“Short cut”
1. Write the fraction with denominator in factored form, and
leave blanks in the numerators of the partial fractions:
2x 1
__
__


( x  3)( x  1) x  3 x  1
2. To get the numerator that goes over x + 3, put your hand
over the x + 3 factor in the fraction on the left and set x = -3 in
the rest. You should end up with 5/2.
3. To get the numerator that goes over x + 1, cover the x + 1 and
set x = -1 (and you get -1/2).
It's that simple.
Another example:
x 1
 x 2  x dx
2
First, the numerator and denominator have
the same degree. So we have to divide it out
before we can do partial fractions.
x 1
x 1
x 1
 1 2  1
2
x x
x x
x( x  1)
2
and we can use partial fractions on the second term
Partial fractions gives:
x 1
A
B
1
 1 
x( x  1)
x x 1
Use either the official or short-cut method
to get A = -1 and B = 2. Therefore:
x2 1
1
2
 x 2  x dx  1  x  x  1 dx  x  ln x  2 ln( x  1)  C
An example for you to try:
3
dx
x (x-1)
Find
2
A) 3/2
E) ln (3/2)
B) 4/3
F) ln (4/3)
C) ln 2
G) ln (2/3)
D) ln 3
H) 3 ln (2)/2
Another example:
4
4x - 6 dx
=
2
x - 3x +2
3
A) ln (4/3)
B) 2 + arctan (3)
C) ln (9)
D) ln (12/5)
E) p/3 - arctan (1/4)
Arc length
The length of a curve in the plane is generally difficult
to compute. To do it, you must add up the little “pieces
of arc”, ds. A good approximation to ds is given by the
2
Pythagorean theorem:
 dy 
ds  dx 2  dy 2  1    dx
 dx 
We can use this to find the length of any graph –
provided we can do the integral that results!
Find the arclength of the parabola y = x2 for x between
-1 and 1. Since dy / dx = 2x, the element of arclength is
2
1 4 x dx
so the total length is:
L
1

1
1  4 x dx
2
So far, we have that the length is
L
1

1  4 x dx
2
1
To do this integral, we will need a trig
substitution. But, appealing to Maple, we
get that 1
L

1
1
1  4t dt  5  ln( 5  2)
2
2
Can we do the integral?
1
The arc length integral from before was:

1  4t 2 dt
1
This is a trig substitution integral of the second kind:
With the identity tan2 + 1 = sec2 in mind, let
What about dt ? Since
t  12 tan 
4t  tan 
2
2
we have that
dt  12 sec 2  d
These substitutions transform the integral into

1  4t dt   sec  d
2
1
2
3
This is a tricky integral we
need to do by parts!
To integrate
Let
sec

d


3
u  sec , dv  sec  d
Then du  sec  tan 
2
, v  tan 
3
2
sec

d


sec

tan


sec

tan
 d


But tan2 = sec2 - 1 , so rewrite the last integral and get
 sec  d  sec tan    sec  d   sec d
3
3
Still going….
 sec  d  sec tan    sec  d   sec d
3
3
It’s remarkable that we’re almost done. The integral
of secant is a known formula, and then you can add
the integral of sec3 to both sides and get
1
1
sec

d


sec

tan


2
2 ln(sec   tan  )  C

3
So we’ve got this so far for 2 1  4t 2 dt where t  12 tan 
We need a triangle!
So far,
with
2 1  4t 2 dt  12 sec tan   12 ln(sec   tan  )  C
t  12 tan 
1  4t 2
From the triangle,
tan   2t , sec  1  4t
So

2t

1
2


1  4t 2 dt  14 1  4t 2 2t   14 ln 1  4t 2  2t  C
Definite integral:
So far, we have



1  4t 2 dt  12 t 1  4t 2  14 ln 1  4t 2  2t  C
Therefore
1

1
1  4 x dx  5 
2
1 ln
4 

5 2

 52
To get the answer Maple got before, we’d
have to rationalize the numerator inside the
logarithm.
Surface Area
The area of a surface of revolution is calculated in a manner
similar to the volume. The following illustration shows the
paraboloid based on y  x (for x=0..2) that we used before,
together with one of the circular bands that sweep out its
surface area.
To calculate the surface area, we first need to determine
To calculate the surface area
the area of the bands. The one centered at the point (x,0)
has radius
x and width equal to ds  dx 2  dy 2.
Since we will be integrating with respect to x (there is a
band for each x), we'll factor the dx out of ds and write
2
 dy 
ds  1    dx . So the area of the band
 dx 
1
dx
centered at (x,0) is equal to: d  2p x 1 
4x
Thus, the total surface area is equal to the integral
1
   p x 4  dx
x
The surface area
turns out to be:
13 p
3
End of methods of
integration!….
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