Mastering the basic Technique of Integration
Today, we are going to learn the efficient method of evaluating indefinite
integrals. This technique will be especially useful, when dealing with
integrals, in the next calculus courses. The technique allows you to
integrate fast, with the minimum use of notations (minimum statements
of the type: "Let u equal to…"), as to avoid the overlap of notations (No
repetition of such statements)
Let's, first recall the ten integration formulas, we know so far.
(Posting the formulas on the board permanently, for the entire
period of the lecture)
These formulas are divided into:
1. The power formula
2. The Logarithm formula (Or the case of a power of -1)
3. The exponential formulas ( 2 formulas)
4. The trigonometric formulas ( six formulas)
1
u n1
1.  u du 
 c ; n  R  {1}
n 1
du
2.  u 1du  
 ln u  c
u
n
3.
u
u
e
du

e
c

au
4.  a du 
c
ln a
u
5.
6.
7.
8.
9.
10.
; a  (0,1)U (1, )
 cos u du  sin u  c
 sin u du   cos u  c
 sec u du  tan u  c
 csc u du   cot u  c
 sec u tan u du  sec u  c
 csc u cot u du   csc u  c
2
2
Introducing the Technique
First: Reminder
Let’s remind ourselves of the method of change of variable, through an
example:
Example (1)
Evaluate:
I = ∫ (x6+2x+2)8 (3x5+1) dx
This is how, we solved this problem
Solution:
2
Let:
u = x6+2x+2
Then:
du = (6x5+2) dx
and so:
dx = du / (6x5+2)
Now:
I = ∫ (x6+2x+2)8 (3x5+1) dx
= ∫ u8 (3x5+1) [du / (6x5+2)]
= ∫ u8 (3x5+1) [du / 2(3x5+1)]
= ∫ u8 (3x5+1) [du / 2(3x5+1)]
=( 1/2) ∫ u8 du
= (1/2)∫ u8 du = ( 1/2) ( u9 / 9 ) + c = (1/18) (3x5+1)9 + c
Second: Justification of the new technique
Let's investigate the meaning of all that!
We concluded at the end of scrubbing all these dummy equations that the
given integral is I is equal to (1/2) ∫ u8 du, and the ∫ u8 is the integral of
(x6+2x+2)8 with respect to (x6+2x+2)!!
This shows that, we could have simply noticed, that:
(x6+2x+2)′ = 6x5+2= 2(3x5+1), which is twice the expression (3x5+1),
and approach the problem, like that:
I = ∫ (x6+2x+2)8 (3x5+1) dx
= ∫ (x6+2x+2)8 (1/2) 2 (3x5+1) dx
= (1/2) ∫ (x6+2x+2)8 (6x5+2) dx
= (1/2) (x6+2x+2)9 / 9 + c = (1/18) (3x5+1)9 + c
Actually, we can omit the second line of the above four lines, and just
write:
I = ∫ (x6+2x+2)8 (3x5+1) dx
= (1/2) ∫ (x6+2x+2)8 (6x5+2) dx
= (1/2) (x6+2x+2)9 / 9 + c
3
Examples (2)
Evaluate:
a. I1 = ∫ cos (x6+2x+2) (3x5+1) dx
b. I2 = ∫ sec2 (x6+2x+2) (3x5+1) dx
c. I3 = ∫ sec(x6+2x+2) tan(x6+2x+2) (3x5+1) dx
d. I4 = ∫ e(x6+2x+2) (3x5+1) dx
e. I5 = ∫ [ 1 / (x6+2x+2)] (3x5+1) dx
Solution:
Using the new technique, we write:
a. I1 = ∫ cos (x6+2x+2) (3x5+1) dx
= (1/2) ∫ cos (x6+2x+2) (6x5+2)dx
= (1/2) sin(x6+2x+2) + c
b. I2 = ∫ sec2 (x6+2x+2) (3x5+1) dx
= (1/2) ∫ sec2 (x6+2x+2) (6x5+2) dx
= (1/2) tan (x6+2x+2) + c
c. I3 = ∫ sec(x6+2x+2) tan(x6+2x+2) (3x5+1) dx
= (1/2) ∫ sec(x6+2x+2) tan(x6+2x+2) (6x5+2) dx
= (1/2) sec(x6+2x+2) + c
d. I4 = ∫ e(x6+2x+2) (3x5+1) dx
= (1/2) ∫ e(x6+2x+2) (6x5+2) dx
= (1/2) e(x6+2x+2) + c
e. I5 = ∫ [ 1 / (x6+2x+2)] (3x5+1) dx
= (1/2) ∫ [ 1 / (x6+2x+2)] (6x5+2) dx
= (1/2) ln |x6+2x+2| + c
4
Understanding the Integration Formulas
The notation u in the ten integration formulas, means any function ☺(x)
of x, while du or, here, d ☺ (the differential of that function☺(x) )
means that we are integrating with respect to u(here with respect to ☺).
Notice that;
d ☺ = ☺′(x) dx
Rewriting the ten formulas, using the "smiling face" notation, which we
will refer to by the " thing"
1.
 (the thing )
n
d (the thing )
(the thing ) n1

 c ; n  R  {1}
n 1
2.
 (the thing )

d (the thing )
 ln (the thing )  c
(the thing )
3.
e
4.
a
5.
 cos(the thing ) d (the thing )  sin( the thing )  c
 sin( the thing ) d (the thing )   cos(the thing )  c
 sec (the thing ) d (the thing )  tan( the thing )  c
 csc (the thing ) d (the thing )   cot(the thing )  c
 sec(the thing ) tan( the thing ) d (the thing )  sec(the thing )  c
 csc(the thing ) cot(the thing ) d (the thing )   csc(the thing )  c
6.
7.
8.
9.
10.
( the thing)
( the thing)
1
d (the thing )
d (the thing )  e ( the thing)  c
a (the thing)
d (the thing ) 
 c ; a is positive , a  1
ln a
2
2
5
Mastering the Basic Technique of integration
Mastering the basic technique of integration means
mastering the two basic skills:
1. The ability to recognize that a certain integral fits one of
the known (in our case: ten) formulas
2. The ability to recognize, that a certain integral can be
made to fit one of the known formulas, by using some simple
algebraic, trigonometric , exponential or logarithmic
manipulation.
The First Skill
The ability to Recognize that a certain integral fits one of the known
(in our case: ten) formulas.
Examples (S-1)
Evaluate:
1. I1 = ∫ cos (lnx) (1 / x ) dx
2. I2 = ∫ sec2 (3x) 3x ln3 dx
3. I3 = ∫ sec(ex) tan(ex) ex dx
4. I4 = ∫ ecosx ( - sinx ) dx
5. I5 = ∫ [ 1 / (secx + 2)] secx tanx dx
6. I6 = ∫ (cosx + lnx + 3)8 [ - sin x + (1/x) ] dx
Solution
Each one of these integrals fits one of the ten integration formula, which
we apply at once
1. I1 = ∫ cos (lnx) (1 / x ) dx
6
We have:
☺ = lnx and d☺ = (1 / x ) dx
This fits the integration formula (5), namely,
∫ cos (☺) ☺′ dx = ∫ cos (☺) d☺ = sin ☺ + c
Thus,
I1 = ∫ cos (lnx) (1 / x ) dx = sin (lnx) + c
2. I2 = ∫ sec2 (3x) 3x ln3 dx
We have:
☺ = 3x and d☺ = 3x ln3 dx
This fits the integration formula (4), namely,
∫ sec2 (☺) ☺′ dx = ∫ sec2 (☺) d☺ = tan ☺ + c
Thus,
I2 = ∫ sec2 (3x) 3x ln3 dx = tan (3x) + c
3. I3 = ∫ sec(ex) tan(ex) ex dx
We have:
☺ = ex and d☺ = ex dx
This fits the integration formula (9), namely,
∫ sec (☺) tan (☺) ☺′ dx = ∫ sec (☺) tan (☺) d☺ = sec ☺ + c
Thus,
I3 = ∫ sec(ex) tan(ex) ex dx= sec ex + c
4. I4 = ∫ ecosx ( - sinx ) dx
We have:
☺ = cosx and d☺ = ( - sinx ) dx
7
This fits the integration formula (9), namely,
∫ e☺ ☺′ dx = ∫ e☺ d☺ = e☺ + c
Thus,
I4 == ∫ ecosx ( - sinx ) dx = ecosx + c
5. I5 = ∫ [ 1 / (secx + 2)] secx tanx dx
= ∫ (secx + 2)-1 secx tanx dx
We have:
☺ = secx + 2 and d☺ = secx tanx dx
This fits the integration formula (2), namely,
∫ ☺-1 ☺′ dx = ∫ ☺-1 d☺ = ln |☺| + c
Thus,
I5 = ∫ [ 1 / (secx + 2)] secx tanx dx = ln | secx + 2 | + c
6. I6 = ∫ (cosx + lnx + 3)8 [ - sin x + (1/x) ] dx
We have:
☺ = cosx + lnx + 3 and d☺ = [ - sinx + (1/x) ] dx
This fits into the integration formulas (1), namely,
∫ ☺n ☺′ dx = ∫ ☺n d☺ = ☺n+1 / (n+1) + c
Thus,
I6 = ∫ (cosx + lnx + 3)8 [ - sin x + (1/x) + 3] dx
= (cosx + lnx + 3)9 / 9 + c
8
Question: Do it now!
Evaluate:
∫ csc2 ( ln lnx ) ( 1 / x lnx) dx
The Second Skill
The ability to recognize, that a certain integral can be made to fit one
of the known formulas, by using some simple algebraic,
trigonometric , exponential or logarithmic manipulation.
Examples (S-2)
Evaluate:
1. I1 = ∫ cos (5lnx) (1 / x ) dx
2. I2 = ∫ sec2 (3x) 3x dx
3. I3 = ∫ csc(5e2x) cot(5e2x) e2x dx
4. I4 = ∫ ecos4x sin4x dx
5. I5 = ∫ [ 1 / (3sec5x + 2)] sec5x tan5x dx
6. I6 = ∫ (sin6x +2 lnx + 3)8 [ 3cos 6x + (1/x) ] dx
Solution
Each one of these integrals can be made to fit one of the ten integration
formulas, which we can apply after doing the required simple
manipulation.
1. I1 = ∫ cos (5lnx) (1 / x ) dx
We have:
9
☺ = 5lnx and d☺ = 5 (1 / x ) dx
We rewrite the integral:
I1 = ∫ cos (5lnx) (1 / x ) dx = (1/5) ∫ cos (5lnx) 5(1 / x ) dx
The integral ∫ cos (5lnx) 5(1 / x ) dx fits the form ∫ cos (☺) d☺
Thus,
I1 = (1/5) sin(5lnx) + c
2. I2 = ∫ sec2 (3x) 3x dx
We have:
☺ = 3x and d☺ = 3x ln3 dx
We rewrite the integral:
I2 = ∫ sec2 (3x) 3x dx = (1/ln3) ∫ sec2 (3x) 3x ln3 dx
The integral ∫ sec2 (3x) 3x ln3 dx fits the form ∫ sec2 (☺) d☺
Thus,
I2 = (1/ln3) tan3x + c
3. I3 = ∫ csc(5e2x) cot(5e2x) e2x dx
We have:
☺ = 5e2x and d☺ = 10e2x dx
We rewrite the integral:
I3 = ∫ csc(5e2x) cot(5e2x) e2x dx
= (1/10) ∫ csc(5e2x) cot(5e2x) 10 e2x dx
The integral ∫ csc(5e2x) cot(5e2x) 10 e2x dx fits the form:
∫ csc (☺) cot (☺) d☺
Thus,
I3 = - (1/10) csc(5e2x) + c
10
4. I4 = ∫ ecos4x sin4x dx
We have:
☺ = cos4x and d☺ = - 4 sin4x dx
We rewrite the integral:
I4 = ∫ ecos4x sin4x dx
= ( - 1 / 4 ) ∫ ecos4x
The integral ∫ ecos4x
( - 4 sin4x )dx
( - 4 sin4x )dx fits the form: ∫ e☺ d☺
Thus,
I4 = - (1/4) ecos4x + c
5. I5 = ∫ [ 1 / (3sec5x + 2)] sec5x tan5x dx
= ∫ (3sec5x + 2)-1 sec5x tan5x dx
We have:
☺ = 3sec5x + 2 and d☺ = 15 sec5x tan5x dx
We rewrite the integral:
I5 = ∫ (3sec5x + 2)-1 sec5x tan5x dx
= ( 1 / 15 ) ∫ (3sec5x + 2)-1 15 sec5x tan5x dx
The integral ∫ (3sec5x + 2)-1 15 sec5x tan5x dx fits the form:
∫ ( 1 /☺ ) d☺
Thus,
I5 = ( 1 / 15 ) ln | 3sec5x + 2 | + c
11
6. I6 = ∫ (sin6x +2 lnx + 3)8 [ 3cos 6x + (1/x) ] dx
We have:
☺ = sin6x +2 lnx + 3 &
d☺ = [ 6cos6x + 2 (1/x) ] dx = 2 [ 3cos6x + (1/x) ] dx
We rewrite the integral:
I6 = ∫ (sin6x +2 lnx + 3)8 [ 3cos 6x + (1/x) ] dx
= ( 1 / 2 ) ∫ (sin6x +2 lnx + 3)8 [ 6cos 6x + 2(1/x) ] dx
The integral ∫ (sin6x +2 lnx + 3)8 [ 6cos 6x + 2(1/x) ] dx fits the form:
∫ ☺n
d☺ with n = 8 ≠ - 1
Thus,
I6 = ( 1 / 15 ) [(sin6x +2 lnx + 3)9 / 9 ] + c
= ( 1 / 135 ) [(sin6x +2 lnx + 3)9 + c
Question: Do it now!
Evaluate:
∫ cos ( 3sin2x) sin2x dx
Hint: sin2x = 2 sinx cosx
&
If ☺ = 3sin2x
Then
d☺ = 6 sinx cosx
12
Problems to Consider at Home
1. Evaluate the following integrals
a. ∫ sin6x cosx dx
b. ∫ tan6x sec2x dx
c. ∫ cot6x csc2x dx
2. Evaluate the following integrals
a. ∫ sin6x cos3x dx
b. ∫ tan6x sec4x dx
c. ∫ cot6x csc4x dx
d. ∫ sec7x tanx dx
e. ∫ (tanx / sec7x) dx
3. Evaluate the following integrals
a. ∫ tanx dx
b. ∫ cotx dx
c. ∫ secx dx
d. ∫ cscx dx
13
4. Evaluate the following integrals
a. ∫ dx / ( x2 +2 x +1)10
b.
∫ dx / ( x3 + 3x2 +3 x +1)10
14