Download Overview Basic Anti-derivatives Simplifying integrals Simple u

Overview
• Familiarize yourself with a list of basic anti-derivatives, and common prototypes.
• Check if the integral is on the basic list, or can be simplified to turn into one(s) on the basic list.
• Check if a simple u-substitution will work.
• Integration by parts: you should have an integral of the product of two functions, or be in a special
case.
• Trigonometric powers
Usually this means you have an integral with powers of sin and cos, or of tan and sec.
• Trigonometric substitution
�
Integrals containing ±x2 ± A2 are usually trigonometric substitution.
• Rational functions:
This means you have an integral of the form
�
polynomial
.
polynomial
• Try again. Some common tricks: rationalizing substutitions which involve square roots, completing
the square within a square root, multiplying by the conjugate.
Basic Anti-derivatives
�
xn+1
x dx =
, if n �= −1
n+1
�
ex dx = ex
�
sin(x) dx = − cos(x)
�
cos(x) dx = sin(x)
�
sec2 (x) dx = tan(x)
�
1
dx = tan−1 (x)
2
� 1+x
1
dx = ln |x|
� x
n
sec(x) tan(x) dx = sec(x)
�
tan(x) dx = ln | sec(x)|
�
√
�
�
�
sec(x) dx = ln | sec(x) + tan(x)|
1
dx = sin−1 (x)
1 − x2
ln(x) dx = x ln |x| − x
1
dx = ln |x + a|
x
+
a
�
1
x
dx =
ln |x2 + a|
2+a
x
2
�
�
�
�x − a�
1
1
�
�
dx =
ln
2
2
2a � x + a �
� x −a
� �
1
1
−1 x
dx
=
tan
x2 + a2
a
a
Simplifying integrals
The most common techniques here are splitting up fractions, for example like this
�
�
�
x
1
ax + b
dx = a
dx + b
dx (∗ can be anything)
∗
∗
∗
and distributing powers, for example like this
�
�
√
(x2 + x − 1) x dx = x5/2 + x3/2 − x1/2 dx
Simple u-substitution
• Identify part of your integral as u,
• Calculate du and see if it contains the rest of the x’s in the integral,
• Translate the original integral into u’s
�
lots of x’s dx =
�
f (u) du
• Find the anti-derivative F (u) and then plug the x’s back into u.
Trigonometric powers overview
• For sine and cosine, the relevent substitutions and identities are
∗ u = sin(x)
∗ u = cos(x)
du = − sin(x) dx
du = cos(x) dx
∗ sin2 (x) = 12 (1 − cos(2x))
cos2 (x) = 12 (1 + cos(2x))
Details for sine and cosine
• For secant and tangent, the relevent substitutions and identities are
∗ u = sec(x)
∗ u = tan(x)
du = sec2 (x)(x) dx
du = sec(x) tan(x) dx
Details for tangent and secant
Integration by parts
When should you use integration by parts?
• Usually: you have an integral with the product of two functions, and the integral would be simpler if
you could replace one of the functions with it’s derivative.
• If you are in a small number of “special cases”, such as
�
�
�
ex sin(x) dx,
ex cos(x) dx,
ln(x) dx,
�
tan−1 (x) dx
How should you do it?
• Product rule notation:
• u and v notation:
�
�
f �g = f g −
u dv = uv −
�
�
f g�
v du
• Notation free notation (fill in the blanks):
�
=
funct funct
anti-deriv orig
−
�
anti-deriv deriv
• Tabular integration by parts
g
−g �
+g ��
−g ���
...
�f
��f
���f
f
...
�
�
��
���
gf = g f − g � f + g �� f − . . .
Trigonometric Substitution overview
�
• Integrals containing ±x2 ± A2 are usually trigonometric substitution. Exception: if the integral has
a single power of x on the outside that makes it work for u-substitution.
• The substitutions in a nutshell:
�
involves
�
2
2
�A − x
2
2
�A + x
2
x − A2
substitution
x = A sin(θ)
x = A tan(θ)
x = A sec(θ)
Rational functions overview
�
polynomial
.
polynomial
You have an integral of the form
• If you can split the fraction up at + and/or − in the numerator and get a basic anti-derivative, then
do this.
• If the degree on top is ≥ the degree on the bottom, then do polynomial division.
• If you can factor the bottom then do partial fractions.
• If you cannot factor the bottom, and the bottom is a quadratic, then complete the square.
Rationalizing substutitions
√
√
If the integral involves
x, try u = x,√along with a small backwards substitution step. More generally,
√
n
n
if the integral involves ax + b try u = ax + b. Note, these substitutions usually involve a backwards
step.
Example
�
Plugging this in gives
To get rid of that last
√
e
√
3x−5
dx
�
√
3x − 5
3
du = √
dx
2 3x − 5
2√
3x − 5 du = dx
3
u=
√
eu 23 3x − 5 du
3x − 5, note that this was the original u-substititon. Thus we get
�
2
ueu du
3
Completing the square within a square root.
Do this to turn a quadratic inside of a square root into a sum or difference of squares:
� �
� �
� �
x2 + 4x + 13 dx =
(x + 2)2 − 9 =
u2 − 9 du
u=x+2
Multiplying by the conjugate
Sometimes it’s best to have more squares:
� �
� �
� �
�
(1 − x)2
1−x
1−x 1−x
1−x
√
√
dx =
·
dx =
dx =
dx
2
1+x
1+x 1−x
1−x
1 − x2
Note that the last thing is much easier to integrate than the first.
Details for
sine and cosine
�
To integrate
sinn (x) cosm (x) dx:
Case 1 The power of sine is odd.
sin2 (x) = 1 − cos2 (x)
u = cos(x)
sin4 (x) = (1 − cos2 (x))2
du = − sin(x) dx
sin6 (x) = (1 − cos2 (x))3 , etc.
Case 2 The power of cosine is odd.
cos2 (x) = 1 − sin2 (x)
u = sin(x)
cos4 (x) = (1 − sin2 (x))2
du = cos(x) dx
cos6 (x) = (1 − sin2 (x))3 , etc
Case 3 Both sine and cosine have even powers.
sin2 (θ) =
1
(1 − cos(2θ))
2
and
cos2 (θ) =
1
(1 + cos(2θ))
2
Multiply everything out, repeat Case 2 for odd powers of cos(2x), repeat Case 3 for even powers of
cos(2x). Evenntually you have only single powers of cos(2x), cos(4x), cos(8x), . . . , which you can finish
immediately.
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Details for
tangent and secant
�
To integrate
tann (x) secm (x) dx:
Case 1 The power of tangent is odd.
u = sec(x)
du = sec(x) tan(x) dx
tan2 (x) = sec2 (x) − 1
tan4 (x) = (sec2 (x) − 1)2
tan6 (x) = (sec2 (x) − 1)3 , etc.
Case 2 The power of secant is even.
sec2 (x) = tan2 (x) + 1
u = tan(x)
du = sec2 (x) dx
sec4 (x) = (tan2 (x) + 1)2
sec6 (x) = (tan2 (x) + 1)3 , etc.
Case 3 Tangent has an even power and secant an odd power.
Get rid of all the powers of tan(x) using tan2 (x) = sec2 (x) − 1 as above. Now we have only powers of
sec(x). Use a integration by parts, the secant-tangent identity, and a little luck.
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Splitting up rational functions
Fractions of numbers can split up:
a+b
a b
= +
c
c
c
and therefore so can rational functions
ax
b
ax + b
= 2
+
cx2 + dx + e
cx + dx + e cx2 + dx + e
This can be useful for integrating:
�
�
�
3x − 5
x
1
1
dx
=
3
dx
−
5
dx = 3 · ln |x2 + 1| − 5 tan−1 (x)
x2 + 1
x2 + 1
x2 + 1
2
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Degrees of polynomials and rational functions
The degree of a polynomial is the biggest power of x that appears. For rational functions, the top and
bottom of the fractions are polynomials, and so the top and bottom have degrees. Partial fractions only
work with the degree of the top being < the degree of the bottom.
Examples:
•
•
4
, degree of top = 0, degree of bottom = 2.
x2 − x + 1
x3 − x + 1
, degree of top =3, degree of bottom = 5 (if you multiply everything out).
x2 (x + 1)3
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Polynomial division
Polynomial division works a lot like regular division:
• at each step, add one more term on top,
• you choose the term so that: when you multiply it by the guy on the side, you can subtract away as
much as possible underneath,
• keep going until the left over stuff underneath has smaller degree than the guy on the side
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Partial fractions
Partial fractions are exactly the same as combining into a common denominator, backwards:
•
Getting common denominator
−−−−−−−−−−−−−−−−−−−−−−−−→
A
B
A(x + 3) + B(x + 4)
+
=
x+2 x+3
(x + 2)(x + 3)
Splitting with partial fractions
• −−−−−−−−−−−−−−−−−−−−−−−−→
4x + 5
A
B
=
+
(x + 2)(x + 3)
x+2 x+3
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Completing the square
Completing the square is a way to turn a general quadratic, x2 + bx + c, into one of the form (x + d)2 + e.
To do this, when the x2 -coefficient is 1 (as it is with x2 + bx + c), take half of b, square this, and add and
subtract the square:
b2
b2
−
+c
x2 + bx +
4
4
Then the first three terms are a perfect square:
�
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x+
�
b 2
2
−
b2
+c
4
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