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1.1. Basic Integration Formulae
Learning Objectives: To evaluate the indefinite integrals by the following seven methods
Making a simplifying substitution
Completing the square
Using a trigonometric identity
Eliminating a square root
Reducing an improper fraction
Separating a fraction
Multiplying by a form of 1 We evaluate an indefinite integral by finding an anti-derivative of the integrand and adding an arbitrary constant. Table 1 shows the basic forms of the integrals we have evaluated so far. Table 1
- (^) ∫ = +
- (^) ∫ = + ( )
- (^) ∫( + ) = ∫ + ∫
- (^) ∫ ^ = ^
+ ^ ( ≠ − 1)
- (^) ∫ = ln| |+
- (^) ∫ = − +
- (^) ∫ = +
- (^) ∫ ^ = +
- (^) ∫ ^ = − +
- (^) ∫ . = +
- (^) ∫ . = − +
- (^) ∫ =
− ln| |+ ln| |+
ln| |+ − ln| |+
14. ∫ ^ = ^ +
15. ∫ ^ = ^
+ ^ ( > 0, ≠ 1)
- (^) ∫ (^) √ (^) = sin ^ +
- (^) ∫ (^) (^) = tan ^ +
- (^) ∫ (^) √ (^) = sec ^ +
- Method of substitution: In this section we reduce certain integrals to some standard forms by a suitable substitution. Example: Evaluate 2
2 9 9 1
x (^) d x x x
Solution:
2 2
1/ 2 1/2 1
1/ 2
2
an 2
2 9 1
x (^) dx x x put u x x du x dx x (^) dx du u du x x u
u (^) C
where C is an arbitrary const t u C
x x C
- Method of completing the square: In this section we reduce certain integrals to some standard forms by completing the square. Example: Evaluate
8 2
dx x x
Solution: We complete the square to write the radicand as
2 2 2
2
2
x x x x x x
x x
x
Then (^8 2 16) 4 ^2 put 4 and 4
dx dx x x (^) x a u x du dx
(^)
2 2 2
1
1
8
sin
tan
sin 4 4
dx du x x a u u (^) C a where C is an arbitrary cons t x (^) C
^
^
x x (^) x x x
Therefore 2
2
2
ln 3 2 3 6 2 3 tan
3 2 ln 3 2 2
x x dx x dx x x dx x dx dx x x x x C
where C is an arbitrary cons t
x x x C
- Method of separating a fraction: In this section we reduce certain integrals to some standard forms by separating a fraction.
Example: Evaluate 2
x
dx
x
Solution: We first separate the integrand to get
2 2 2
x x dx
dx dx
x x x
In the first integral we substitute u 1 x 2 du 2 x d x
tan
x du
dx
x u
u du
u
C x C
whereC is an arbitrary cons t
The second integral is in the standard form 1 2 2
2
2 2 sin 1 ta n
d x (^) x C x w h e re C is a n a rb itra ry co n s t
Combing these results and renaming C 1 C 2 as C gives
2 1 2
3 1 2 sin 1 tan
x d x x x C x w h ere C is an a rbitra ry con s t
= ln| + |+
- Method of multiplying by a form of 1: In this section we reduce certain integrals to some standard forms by multiplying by a form of 1.
Example: Evaluate (^) sec x dx
Solution:
2
2
sec (sec ) 1 sec tan sec sec tan sec sec tan sec tan Substitute sec tan sec tan sec
sec
ln ln sec tan tan
x dx x dx x x x dx x x x x x dx x x u x x du x x x dx
x dx du u u C x x C where C is an arbitrary cons t
With cosecants and cotangents in place of secants and tangents, the method of the above example leads to a companion formula for the integral of the cosecant.
PROBLEM SET
IP1. Evaluate: ∫
^ ^ ^?
Solution:
Given (^) ∫ (^) = ∫ √ (^)
(^)
Now, put
(^) = ⟹ √ =
∫ √ (^)
=^
∫^
(^)
= sin ^
+ =^
sin
^
where C is an arbitrary constant
∴ ∫ (^) = sin ^
(^) +
= − ln| + |+
^ = ∫^
√
= sin ^
√
+ = sin ^
where C is an arbitrary constant
= sin ^
√ + = sin
^ = sin
√ ^ +
IP3. Evaluate: ∫
Solution:
√ = ∫^
( )
√ ^
√ ^
= ∫ ^ + ^ + 2 . − ∫ √
.
√
.
√
. .
= 2 − + (^) √ ln + + + + where C is an arbitrary constant
∴
ln
P3. Evaluate: ∫
.
( ) ^?
Solution:
.
( ) =^
∫^
.
∫^
.
∫^
^ ^ .
∫^
( ) ^
∫( + ) −^
∫^
∫( + ) −^
∫^
√ √ ^
(− + ) −^
√ ∫^
√ ^ ^
√ ^ ^
√ ∫^
( − ) −^
√ ∫^
( − ) −^
√ ∫ +^
( − ) +^
√ ln +^
+ +^
where C is an arbitrary constant
ln +
IP4. Evaluate: (^) ∫ ^
(^) ? Solution:
Given (^) ∫ ^
(^) ^ =^
( ) ^ = ( − 2) −^
( ) ∴ ∫ ^
(^) = ∫ ( − 2) −^
( ) = ∫ − 2 ∫ − 2 ∫ (^) ( )
= ^
− 2 − 2 ln| − 2|+ where C is an arbitrary constant ∴ ∫ ^
(^) =^
− 2 − 2 ln| − 2|+
P4. Evaluate: (^) ∫ ^
(^) ? Solution:
Given (^) ∫ ^
(^) By actual division, we have ^ =
−^
+^
( ) ∴ ∫ ^
(^) = ∫
−^
+^
( ) = ∫ − ∫ + ∫ (^) ( )
= ^
−^
+^
| | + where C is an arbitrary constant = ^
−^
+^
ln|2 + 3|+
) ∫ ^
(^) (^) ^ ^ ) ∫ ^
^ ) ∫ ^
√ ^
- Evaluate
)
^ + 2
^ + 1
) ∫ ^
√ ^ ( √ ^ ) ) ∫ ^
^
1.2. Integration by parts
Learning objectives: To derive the formula for integration by parts. To execute the integration by parts by tabular integration. And To practice related problems.
Introduction: Integration by parts is a technique for simplifying integrals of the form
f^ ^ x^ ^ g^ ^ x dx
in which f ( x) can be differentiated repeatedly and g x( )can be integrated repeatedly
without difficulty.
The integral (^) xe dxx
is such an integral because f (^) x (^) xcan be differentiated twice to become zero and
g (^) x (^) ex can be integrated repeatedly without difficulty.
Integration by parts also applies to integrals like
e^ x^ sinx dx in which each part of the integrand appears again after repeated differentiation or integration. The Formula : The formula for integration by parts comes from the product rule,
d d v d u u v u v d x d x d x
In its differential form, the rule becomes
d uv u d v v d u
which is then written as u dv^ ^ d uv ^ vdu and integrated to give the following
formula.
u dv^ ^ uv^ v du .................. (1)
Equation(1) is the integration-by-parts formula and it expresses one integral, (^) u dv, in
terms of a second integral, (^) v du.
With a proper choice of u and v , the second integral may be easier to evaluate than the first. (This is the reason for the importance of the formula. When faced with an integral we cannot handle, we can replace it by one with which we might have more success) The equivalent formula for definite integrals is
2 2 1 1
v u v u u dv^ ^ u v^ ^ u v^ v du
Example-1: Find (^) x cosx dx
Solution: Let co s cos sin
(^)
u x and dv x d x d u d x and v x dx x
u dv^ ^ u v^ v du
co s sin sin
sin co s
x^ x d x^ x^ x^ x d x
x x x C
where C is an arbitrary constant.
Example-2: Find (^) ln x dx.
Solution:
Since (^) ln x dxcan be written as (^) ln x 1 dx, we use the
formula (^) u dv uv v duwith
ln
(^)
u x a n d d v d x
d u d x a n d v d x x
x
Then (^) u dv uv v du
ln ln
ln
x dx x x x dx
x
x x x C
where C is an arbitrary constant
This gives u dv uv v du
x e dx^2 x^ x e^2 x^ 2 xe dxx ....(1)
It takes a second integration by parts to find the integral on the right. We find x
x x
u x and dv e dx
du dx and v e dx e
u dv^ ^ uv^ v du
xe dx^ x^ xe x^ e xC
,^ where^ ʹ^ is an arbitrary constant
from (1)
x e dx^2 x^ x e^2 x^ 2 xe dxx ....(1)
x e dx^2 x^ x e^2 x^ 2 xe x^ ex
Hence x e dx^2 x^ x e^2 x^ 2 xe x^ 2 e x C
where is an arbitrary constant Solving for the Unknown Integral Integrals like the one in the next example occur in electrical engineering. Their evaluation requires two integrations by parts, followed by solving for the unknown integral.
Example: Find e xcosx dx
Solution: We first use the formula u d v u v v d u with
cos
cos sin
x
x
u e and dv x dx
du e dx and v x dx x
Then
e^ x^ cos^ x dx^ ^ e^ x^ sin^ x^ e^ xsinx dx
The second integral is like the first, except it has sin x in place of cos x. To evaluate it, we use integration by parts with
sin
sin cos
x
x
u e and dv x dx
du e dx and v x dx x
Then
cos sin cos cos
sin cos cos
x x x x
x x x
e x dx e x e x x e dx
e x e x e x dx
The unknown integral now appears on both sides of the equation. Combining the two expressions gives
(^2) e x^ co s x d x e x^ sin x e xcosx C where ʹ is an arbitrary constant Dividing by 2 and renaming the constant of integration gives
sin cos
cos
x x
e x^ x dx e^ x^ e^ xC
where is an arbitrary constant Tabular Integration
We have seen that integrals of the form (^) f (^) x (^) g (^) x (^) d x, in which f can be
differentiated repeatedly to become zero and g can be integrated repeatedly without
difficulty, are natural candidates for integration by parts. In some cases, where there are many repetitions, the calculations can be cumbersome. There is a way to organize the calculations that saves a great deal of work. It is called tabular integration and is illustrated in the following examples.
Example : Find (^) x 2 e xd x by tabular integration.
Solution: With f (^) x (^) x^2 and g (^) x (^) ex, we list
We add the products of the functions connected by the arrows, with the middle sign
changed, to obtain (^) x 2 e x^ dx x 2 e x^ 2 xe x^ 2 e x C
where C is an arbitrary constant
Example: Find (^) x 3 sinx dx by tabular integration.
Solution: With f (^) x (^) x^3 and g x sinx, we list
IP2. Evaluate: (^) ∫ ^ ? Solution: Given (^) ∫ ^ 2 Put = ^ and = 2 ⟹ = − ^ and =
∫ ^ ^ 2 =^
^ − ∫^
(− ^
= ^
(^) +^
∫ ^
Now, to find (^) ∫ ^ 2 Put = ^ and = 2 ⟹ = − ^ and = ∫ 2 = − By integration by parts, we have ∫ = − ∫
∫ ^ ^ 2 = −^
^ + ∫^
(− ^
= − ^
(^) −^
∫ ^
From (1), we have ∴ ∫ ^ 2 = ^
(^) +^
−^
^ −^
∫ ^
= ^
(^) −^
^ −^
∫ ^
1 + ∫ ^ 2 = ^
(2 2 − 2 ) + ʹ ∫ ^ ^ 2 =^
(2 2 − 2 ) + where is an arbitrary constant P2. Evaluate: (^) ∫ ^ ? Solution: Given (^) ∫ ^ 2 Put = ^ and = 2 ⟹ = − ^ and = ∫ 2 = − By integration by parts, we have ∫ = − ∫ ∫ ^ ^ 2 = −^
^ − ∫ −^
(− ^
= − ^
(^) −^
∫ ^
Now, to find (^) ∫ ^ 2 Put = ^ and = 2 ⟹ = − ^ and = ∫ 2 = By integration by parts, we have ∫ = − ∫ ∫ ^ ^ 2 =^
^ − ∫^
(− ^
= ^
(^) +^
∫ ^
From (1), we have
∴ ∫ ^ 2 = − ^
(^) −^
∫ ^
= − ^
(^) −^
^ +^
∫ ^
= − ^
(^) −^
^ −^
∫ ^
∫ ^ ^ 2 +^
∫ ^
2 = − ^ ^
−^
^ 1 + ∫ ^ 2 = − ^
(2 2 + 2 ) + ʹ ∫ ^ ^ 2 = −^
(2 2 + 2 ) + where is an arbitrary constant IP3. Evaluate: (^) ∫ ( + ) ? Solution: Given (^) ∫ log( + 1) = and = log( + 1) ⟹ = and = ∫ log( + 1) = ( + 1)[log( + 1) − 1] By integration by parts, we have ∫ = − ∫ ∫ log( + 1) = ( + 1)[log( + 1) − 1]− ∫( + 1)[log( + 1) − 1] = ( + 1)[log( + 1) − 1]− ∫( + 1) log( + 1) + ∫( + 1) = ( ^ + )[log( + 1) − 1]− ∫ log( + 1) − ∫ log( + 1) + ^
+ 2 ∫ log( + 1) = ( ^ + )[log( + 1) − 1]− ( + 1)[log( + 1) − 1]+ ^
+ + = ( ^ + − − 1)[log( + 1) − 1]+ ^
+ + = ( ^ − 1)[log( + 1) − 1]+ ^
+ + = ( ^ − 1) log( + 1) − ^ + 1 + ^
+ + ∫ log( + 1) =^
( ^
(^) − 1) log( + 1) − ^ + + 1 + where is an arbitrary constant P3. Evaluate: (^) ∫ √ ? Solution: Given (^) ∫ √ log Put = √ and = log ⟹ = (^) √ and = (log − 1) By integration by parts, we have ∫ = − ∫ ∫ √ log = √ (log − 1) − ∫ (log − 1)^
√ =
(^) (log − 1) − ∫ √ (log − 1)
By tabular integration,
∴ ^ 2
= ^
(^) − 3 ^
+ 6 −^
− 6
+ where C is an arbitrary constant = ^
(^) +^
^ −^
−^
+
EXERCISES
1. Evaluate
a. (^) ∫ sin ^ b. (^) ∫ c. (^) ∫ d. (^) ∫ tan ^ e. (^) ∫ log(1 + ^ ) f. (^) ∫ ^ log g. (^) ∫ sin ^ h. (^) ∫ cos ^ i. (^) ∫ tan ^ j. (^) ∫ log + √ ^ + ^
2. Evaluate
a. (^) ∫ ^ 4 b. (^) ∫ ^ 4 c. (^) ∫ ^ 3 d. (^) ∫ ^ log(1 + ) e. (^) ∫ ^ log
3. Evaluate by tabular integration
a. (^) ∫ ^ ^ b. (^) ∫ ^ ^ c. (^) ∫( ^ + 2 ^ + 5 ^ + + 1) ^ ^ d. (^) ∫ ^ (log ) ^ e. (^) ∫ ^ 2 f. (^) ∫ ^ 2
1.3. Partial Fractions
Learning objectives: To express the rational fractions into simpler fractions by the method of partial fractions. To evaluate the integrals by the method of partial fractions in which the integrand consists of Distinct linear factors in the denominator. A repeated linear factor in the denominator. An irreducible quadratic factor in the denominator. AND To practice the related problems.
A theorem from advanced algebra says that every rational function, no matter how complicated, can be rewritten as a sum of simpler fractions that we can integrate with techniques we already know. For instance, 5 3 2 3 (^2 2 31 )
x x x x^ x
^
So, we can integrate the rational function on the left by integrating the fractions on the right instead. The method for rewriting rational functions in this way is called the method of partial
fractions. In this particular case, it consists of finding constants A and B such that
5 3 5 3 (^2 2 31 3 1 )
x x A B x x x^ x^ x^ x
(^) ^ ^ ^
We call the fractions A / (^) x (^1) and B / (^) x (^3) as partial fractions because their
denominators are only part of the original denominator (^) x 2 2 x 3. We call A and B undetermined coefficients until proper values for them have been found.
To find A and B, we first clear the equation (2) of fractions, obtaining
5 x 3 A (^) x (^3) B (^) x (^1) (^) A B (^) x 3 A B
This will be an identity in x if and only if the coefficients of like powers of x on the two sides are equal: (^) A B 5 , 3 A B 3
Solving these equations simultaneously gives A 2 and B 3. Example 1- Two distinct linear factors in the denominator Find :
5 3 1 3
x (^) d x x x
Solution:
5 3 2 3 1 3 1 3
x x x x x
(^)
5 3 2 3 1 3 1 3 2 ln 1 3 ln 3
x (^) d x d x d x x x x x x x C
(^)
where C is an arbitrary constant
