Download Review of Integration Techniques

REFRESHER 2: INTEGRATION TECHNIQUES
MA26200, Gabriel Sosa
June 23, 2014 - Monday
1. COMPOSITION WITH LINEAR FUNCTIONS
Integration is the opposite of differentiation.
R
For example cos xdx = sin x + C, since (sin x + C)0 = cos x.
Functions that are the derivative of a known function are easy to
integrate. For example:
Z
Z
1
(sec x tan x)dx = sec x + C
dx = ln |x| + C
x
Z
Z
1
(sec2 x)dx = tan x + C
dx = arctan x + C
1 + x2
We also possess the power rule for integrals
Z
xn+1
n
x dx =
+ C whenever n 6= −1.
n+1
Z
Z
1
x−2
1
−3
So
dx
=
x
dx
=
+
C
=
−
+ C.
x3
−2
2x2
This also applies when we want to integrate:
Z 2
Z
Z
1
x5/2 x1/2
x +1
x2
3/2
−1/2
√ dx =
+
dx
=
x
+
x
dx
=
+
+C
5/2 1/2
x
x1/2 x1/2
√
2√ 5
=
x +2 x+C
5
Sometimes you will have a known derivative composed with a linear
function. In those cases integrate and divide by the coefficient of x
in
Z the linear function:
cos (2x + 3)
sin (2x + 3)dx = −
+C
2
Z
1
ln |3−5x|
dx =
+C
3−5x
−5
Refresher 2
MA26200 (Gabriel Sosa), Summer ’14
Page 1 of 12
2. THE SUBSTITUTION METHOD
Question: When could substitution work?
Answer: When the function to be integrated is one of the following:
1. A division. In this case try the denominator as the substitution.
For example:
Z
Z
cos x
1
u = sin x − 3
•
dx =
cos xdx. So
du = cos xdx
sin x − 3
sin x − 3
Z
Z
1
cos x
dx =
du = ln |u| + C = ln | sin x − 3| + C
sin x − 3
u
Z
Z
1
e2x
u = 1 + ex
2x
. Notice ex = u−1
dx =
e dx. So
•
x
x
x
du = e dx
1+e
1+e
Z
Z
Z
Z
e2x
1
1
1
x x
dx
=
e
e
dx
=
(u
−
1)du
=
(1
−
)du
1 + ex
1 + ex
u
u
Z
e2x
dx = u − ln u + C = 1 + ex − ln (1 + ex ) + C
x
1+e
Z
e2x
dx = ex − ln (1 + ex ) + C. The constant absorbed the 1.
1 + ex
In rare occasions, writing the division as a product is better and we
apply:
2. A product. If we have the product of two functions where one
is the derivative of the other. For example:
Z
Z
ln x
1
1
u = ln x
•
dx = ln x dx. The derivative of ln x is , so
du = x1 dx
x
x
x
Z
Z
Z
ln x
1
u2
(ln x)2
dx = ln x
dx = udu =
+C =
+C
x
x
2
2
R
• sec2 x tan xdx. The derivative of tan x is sec2 x, so u = tan x
2
and
Z du = sec xdx Z
Z
u2
2
2
sec x tan xdx = tan x sec xdx = udu =
+C
2
Z
tan2 x
2
So sec x tan xdx =
+C
2
Refresher 2
MA26200 (Gabriel Sosa), Summer ’14
Page 2 of 12
3. A composition When there is one composition present. In this
case, try the inner function being composed as the substitution.
For example:
u = ex
• e sin e dx. Since sin e is a composition, we take
du = ex dx
R
R
R x
e sin ex dx = (sin ex )ex dx = (sin u)du = − cos u + C
R
Hence ex sin ex dx = − cos ex + C
R
Z
•
x
ecos
x
2
x
x
cos x sin xdx. Since ecos
2
x
is a composition, we consider
du
u = cos2 x and du = 2 cos x(− sin x)dx, or,
= cos x sin xdx
−2
Z
Z
−1 u
1 2
du
2
=
e + C = − ecos x + C.
ecos x cos x sin xdx = eu
−2
2
2
Z
u = sin x
•
sin3 x cos xdx. Since sin3 x is a composition,
du = cos xdx
Z
Z
4
u4
sin x
sin3 x cos xdx = u3 du =
+C =
+C
4
4
Z
p
√
u = x2 + 1
.
•
x3 x2 + 1dx. Since x2 + 1 is a composition,
du = 2xdx
du
So we have u = x2 + 1,
= xdx and x2 = u − 1.
Z
Z 2p
Z
p
√
du
3
2
So x x2 + 1dx =
x2 + 1x xdx =
u(u − 1)
2
Z
Z
p
1
1 u5/2 u3/2
x3 x2 + 1dx =
−
u3/2 − u1/2 du =
+C
2
2 5/2
3/2
Z
p
1
2
1
1
2
3
5/2
3/2
x x2 + 1dx =
u − u
+ C = u5/2 − u3/2 + C
2 5
3
5
3
Z
p
3
1p 2
1p 2
x3 x2 + 1dx =
(x + 1)5 −
(x + 1) + C.
5
3
Refresher 2
MA26200 (Gabriel Sosa), Summer ’14
Page 3 of 12
3. INTEGRATION BY PARTS
Question: When do we apply integration by parts?
Answer: When we have a product of BASIC function whose derivatives
are unrelated. (By BASIC I mean no compositions are present).
Question: How do we apply it?
Answer: Using the rule:
You destroy vines, you vineman, less vines destroy you
Z
Z
udv = uv −
vdu
along with the method to identify u-Log Alg Trig Exp
Examples
R
(a) x3 ln xdx, by u-LATE, we get u = ln x and dv = x3 dx.
Z 4
Z
4
4
x 1
x
u = ln x
v = x4
dx.
, and x3 ln xdx = ln x −
So
1
3
du = x dx dv = x dx
4
4x
Z
Z
x4 ln x 1
x4 ln x 1 x4
x4 ln x x4
3
3
x ln xdx =
x dx =
−
−
+C =
− +C
4
4
4
44
4
16
(b)
R
x sin xdx, by u-LATE, we get u = x and dv = sin xdx.
R
R
u=x
v = − cos x
So
, and x sin xdx = x(− cos x)− (− cos x)dx
du = dx dv = sin xdcx
R
R
x sin xdx = −x cos x + cos xdx = −x cos x + sin x + C.
Refresher 2
MA26200 (Gabriel Sosa), Summer ’14
Page 4 of 12
Sometimes we need to apply the method more than once
R
x2 ex dx, by u-LATE we get u = x2 and dv = ex dx.
R 2 x
R x
u = x2
v = ex
2 x
So
,
and
x
e
dx
=
x
e
−
e 2xdx
du = 2xdx dv = ex dx
R 2 x
R
x e dx = x2 ex − 2 xex dx, so we have to do it again.
R
xex dx, by u-LATE, we get u = x and dv = ex dx
R x
R x
u=x
v = ex
x
So
,
and
xe
dx
=
xe
−
e dx = xex − ex
du = dx dv = ex dx
R
R
x2 ex dx = x2 ex − 2 xex dx = x2 ex − 2 [xex − ex ] + C
R
So x2 ex dx = x2 ex − 2xex + 2ex + C
And, sometimes, we need to apply the method twice and be clever
R
ex cos xdx, by u-LATE we get u = cos x and dv = ex dx
u = cos x
v = ex
, and
So
du = − sin xdx dv = ex dx
R x
R
R
e cos xdx = cos xex − ex (− sin x)dx = ex cos x + ex sin xdx, so
we have to do it again:
R
ex sin xdx, by u-LATE we get u = sin x and dv = ex dx
R x
R x
u = sin x
v = ex
x
So
,
e
sin
xdx
=
sin
xe
−
e cos xdx
du = cos xdx dv = ex dx
R
R
Hence ex cos xdx = ex cos x + ex sin x− ex cos xdx
R
R
So ex cos xdx + ex cos xdx = ex cos x + ex sin x
R
And 2 ex cos xdx = ex cos x + ex sin x, which means
Z
ex (cos x + sin x)
x
e cos xdx =
+C
2
Refresher 2
MA26200 (Gabriel Sosa), Summer ’14
Page 5 of 12
4. TRIGONOMETRIC SUBSTITUTION
Question: When do we apply trigonometric substitution?
Answer: When an expression involving an addition or a subtraction of
squares appears AND substitution does not work).
Z
Z
p
x
3
For example, the integrals
x 1 − x2 dx and
dx should be
x2 + 4
solved by substitution. [Check at the end of this section for details].
Examples:
(a)
Z
1
dx
x x2 + 4
√
In this case we have x2 +22 ; since
we have an addition, we will create
a right triangle with legs x and 2
We need to relate the
angle θ to x:
x
tan θ =
2
So, 2 tan θ = x and
On the other hand, we
also need to
√ relate the
angle θ to x2 + 4:
√
x2 + 4
sec θ =
2
√
2 sec2 θdθ = dx
So, 2 sec θ = x2 + 4.
Z
Z
Z
1
sec θ
1
2
√
dx =
2 sec θdθ =
dθ
Hence
2 tan θ2 sec θ
2 tan θ
x x2 + 4
Z
Z 1
Z
Z
1
1
1
1
1
cos
θ
√
dx =
dθ =
dθ =
csc θdθ
sin θ
2
2
sin
θ
2
x x2 + 4
cosθ
Z
1
1
√
dx = ln | csc θ − cot θ| + C
2
x x2 + 4
√
x2 + 4
2
Finally from the triangle we know that csc θ =
, while cot θ = .
x √
x
√
Z
1
1 x2 + 4 2 1 x2 + 4 − 2 √
− + C = ln So
dx = ln +C
2 x
x
2 x
x x2 + 4
Refresher 2
MA26200 (Gabriel Sosa), Summer ’14
Page 6 of 12
(b)
Z p
9 − x2 dx
In this case we have 32 −x2 ; since
we have a subtraction, we will
create a right triangle with
hypotenuse 3 and leg x
We need to relate the On the other hand, we
angle θ to x:
also need to
√ relate the
angle θ to 9 − x2 :
x
sin θ =
√
3
9 − x2
cos
θ
=
So, 3 sin θ = x and
3
√
3 cos θdθ = dx
So, 3 cos θ = 9 − x2 .
Z
Z p
Z
Z
2
Hence
9 − x2 dx = 3 cos θ3 cos θdθ = 9 cos θdθ = 9 cos2 θdθ
The key to integrating cos2 θ lies on one of the identities in the Trigonometric Identitites Refresher :
cos 2θ + 1
.
cos 2θ = 2 cos2 θ − 1 or its equivalent cos2 θ =
2
Z p
Z
9
9
sin
2θ
So
9 − x2 dx =
(cos 2θ + 1)dθ =
+θ +C
2
2
2
Now remember sin 2θ = 2 sin θ cos θ, so:
Z p
2
sin
θ
cos
θ
9
9θ
9
+ θ + C = sin θ cos θ +
+C
9 − x2 dx =
2
2
2
2
√
x
9 − x2
Finally from the triangle we know that sin θ = , while cos θ =
.
3
3
√
Z p
x
x 9 − x2
9
9
2
So
9 − x dx =
+ arcsin
+C
2 3
3
2
3
√
Z p
x
x
9 − x2 9
2
And we conclude
9 − x dx =
+ arcsin
+C
2
2
3
Refresher 2
MA26200 (Gabriel Sosa), Summer ’14
Page 7 of 12
(c)
Z
√
1
x2 − 1
In this case we have x2 −12 ; since
we have a subtraction, we will
create a right triangle with
hypotenuse x and leg 1
dx
We need to relate the On the other hand, we
also need to
angle θ to x:
√ relate the
angle θ to x2 − 1:
csc θ = x
√
cot θ = x2 − 1
− csc θ cot θdθ = dx
Z
1
Z
1
cot
θ)dθ = −
(− csc θ
cot θ
√
Hence
dx =
x2 − 1
Z
1
So √
dx = − ln |csc θ − cot θ| + C
x2 − 1
Z
csc θdθ
Finally from the triangle we know that csc θ = x, while cot θ =
Z
p
1
2
√
dx = − ln x − x − 1 + C
And we conclude
x2 − 1
p
x2 − 1.
As for the integrals discussed at the beginning of the section
R 3√
x 1 − x2 dx. This example shows that before trying trigonometric
substitution one needs to check that substitution does not work.
√
u = 1 − x2
Since 1 − x2 is a composition, we should try
, so we
du = −2xdx
du
have u = 1 − x2 ,
= xdx and x2 = 1 − u.
−2
Z
Z
Z
p
p
√ du
x3 1 − x2 dx = x2 1 − x2 xdx = (1 − u) u
−2
Z
Z
3/2
5/2
p
1
1
u
u
x3 1 − x2 dx = −
u1/2 − u3/2 du = −
−
+C
2
2 3/2
5/2
1
1
1 2 3/2 2 5/2
−
u − u
+ C = − (1 − x2 )3/2 + (1 − x2 )5/2 + C
2 3
5
3
5
Refresher 2
MA26200 (Gabriel Sosa), Summer ’14
Page 8 of 12
Z
x
dx, this is a division, so we should first try the substitution
x2 + 4
du
u = x2 + 4, du = 2xdx, or equivalently
= xdx.
2
Z
Z
Z
Z
1
1 du 1
1
x
1
dx
=
xdx
=
=
du
=
ln |u| + C
x2 + 4
x2 + 4
u 2
2 u
2
Z
p
x
1
2
dx = ln |x + 4| + C = ln x2 + 4 + C
x2 + 4
2
Notice that |x2 + 4| = x2 + 4, because x2 + 4 is always positive.
Remark: You should have already memorized that
Z
x
1
1
dx = arctan
+C
x2 + a2
a
a
Refresher 2
MA26200 (Gabriel Sosa), Summer ’14
Page 9 of 12
5. PARTIAL FRACTIONS
Question: When do we use partial fractions?
Answer: When we have a division of polynomials, the degree of the
numerator is LESS than the degree
Z of the denominators AND substitux−2
tion does not work. For example
dx should be integrated
x2 − 4x − 5
using substitution, not partial fractions.
Question: How do we find the partial fraction decomposition
of a rational expression?
Answer: Let’s look at the following two examples
x−9
(a) 2
. The expression provided by:
(x + 4)(x + 1)
Irreducible quadratic factors x2 + 4 are of the form
Ax + B
.
x2 + 4
C
x+1
Ax + B
C
x−9
What we want then is 2
+
= 2
x +4
x + 1 (x + 4)(x + 1)
So (Ax + B)(x + 1) + C(x2 + 4) = x − 9. There are two ways of finding
the values for A, B and C.
Linear factors x + 1 are of the form
Solution 1:
If x = −1
If x = 0
If x = 1
then 5C = −10,
then B + 4C = −9,
(A + B)2 + 5C = −8,
and C = −2
B−8 = −9, so B = −1 2(A − 1) − 10 = −8,
2A − 12 = −8, A = 2
Solution 2:
(Ax + B)(x + 1) + C(x2 + 4) = x − 9
Ax2 + Ax + Bx + B + Cx2 + 4C = x − 9
(A + C)x2 + (A + B)x + (B + 4C) = x − 9
So A + C = 0, A + B = 1 and B + 4C = −9
Then C = −A, B = 1 − A and 1 − A − 4A = −9
So 5A = 10, and A = 2, B = −1 and C = −2
Refresher 2
MA26200 (Gabriel Sosa), Summer ’14
Page 10 of 12
2x − 1
2
x−9
=
−
,
(x2 + 4)(x + 1)
x2 + 4 x + 1
Z
Z
2
x−9
2x − 1
and then
dx =
−
dx
2
2
(x + 4)(x + 1)
x +4 x+1
Z
Z
Z
2x
1
2
=
dx
−
dx
−
dx
x2 + 4
x2 + 4
x+1
Z
x
1
x−9
2
dx = ln (x + 4) − arctan
− 2 ln |x + 1| + C
So
(x2 + 4)(x + 1)
2
2
All these integrals should be known.
Hence
x2 − 2x + 3
(b)
. The expression provided by:
(x − 1)2 (x − 2)
A
Linear factors x − 2 are of the form
x−2
Powers of linear factors (x − 1)2 are of the form
B
C
+
x − 1 (x − 1)2
A
B
C
x2 − 2x + 3
So
+
+
=
and
x − 2 x − 1 (x − 1)2
(x − 1)2 (x − 2)
A(x − 1)2 + B(x − 2)(x − 1) + C(x − 2) = x2 − 2x + 3
Once away there are two ways to find the values of A, B and C:
Solution 1:
If x = 1,
If x = 2,
If x = 0, then
then −C = 2,
then A = 3
A + 2B − 2C = 3
so C = −2
so 3 + 2B + 4 = 3,
and B = −2
Solution 2:
A(x2 − 2x + 1) + B(x2 − 3x + 2) + Cx − 2C = x2 − 2x + 3
Ax2 − 2Ax + A + Bx2 − 3Bx + 2B + Cx − 2C = x2 − 2x + 3
(A + B)x2 + (C − 2A − 3B)x + (A + 2B − 2C) = x2 − 2x + 3
So A + B = 1, C − 2A − 3B = −2 and A + 2B − 2C = 3
Then B = 1−A, C −2A−3(1−A) = −2 or C −2A−3+3A = −2, which
shows C = 1 − A. then B = C. We conclude A = 3 and B = C = −2
Refresher 2
MA26200 (Gabriel Sosa), Summer ’14
Page 11 of 12
x2 − 2x + 3
So the partial fractions representation for
is
(x − 1)2 (x − 2)
2
2
3
−
−
.
x − 2 x − 1 (x − 1)2
Z
Z
x2 − 2x + 3
3
2
2
This implies that
dx
=
−
−
dx
(x − 1)2 (x − 2)
x − 2 x − 1 (x − 1)2
Z
Z
Z
1
1
=3
dx − 2
dx − 2 (x − 1)−2 dx
x−2
x−1
|x − 2|3
2
= 3 ln |x − 2| − 2 ln |x − 1| + 2(x − 1)−1 + C = ln
+
+C
|x − 1|2 x − 1
One last persistent note:
ALWAYS TRY SUBSTITUTION FIRST when integrating divisions and/or
compositions. If that does not work you can try TRIGONOMETRIC SUBSTITUTION or PARTIAL FRACTIONS.
INTEGRATION BY PARTS should only be applied to PRODUCTS of
basic functions.
To
Z stress my point:
x−2
dx should be solved by substituting u = x + 3. I invite you to
(x + 3)2
try this byZ substitution and by partial fractions so you can see the difference.
p
While
x 4 − x2 dx should be solved by substituting u = 4 − x2 . Try
it by substitution and then by trigonometric substitution so you can tell the
difference
And remember, all the material in this refresher covers things
you were supposed to have learned already. Hopefully this taught
you a lot about how cumulative knowledge works.
Refresher 2
MA26200 (Gabriel Sosa), Summer ’14
Page 12 of 12