IMPORTANT FORMULAS IN INTEGRATION
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Introduction and Definition of Integration
Consider a function 𝑓(𝑥). When we differentiate it with respect to 𝑥, we get:
𝑑
𝑑𝑥
[𝑓 (𝑥 )] = 𝑓 ′ (𝑥) ; where 𝑓 ′ (𝑥) is the derivative of 𝑓(𝑥 ).
Now, when we integrate 𝑓 ′ (𝑥), i.e.
∫ 𝑓 ′ (𝑥) 𝑑𝑥 = 𝑓 (𝑥 ) + 𝑐; where 𝑓 (𝑥 ) is called the integral of 𝑓 ′ (𝑥) and c is the
constant of integration. The operator 𝑑𝑥 indicates that we are integrating w.r.t. 𝑥.
We need the constant of integration because the integral of any term is not definite
i.e. the constant lost in differentiation will not be recovered on integration.
For example, if we differentiate a function 𝑓(𝑥 ) = 3𝑥 2 + 5𝑥 + 7, we get 𝑓 ′ (𝑥 ) =
6𝑥 + 5. Now, while integrating 𝑓 ′ (𝑥), we get ∫ 𝑓 ′ (𝑥) 𝑑𝑥 = ∫(6𝑥 + 5)𝑑𝑥 which will
be equal to 3𝑥 2 + 5𝑥. In other words, the constant ‘7’ which is lost during the
differentiation process is not retrieved in the integration process. To compensate this,
we add an arbitrary constant c known as constant of integration.
The process of determining an integral of a function is called integration and the
function to be integrated is called the integrand.
Since the integral of any function is not completely defined, it is known as Indefinite
Integration.
Therefore, we see that integration is the reverse process of differentiation.
Basic Integrals of Algebraic Functions
1.
2.
𝑛
∫ 𝑥 𝑑𝑥 =
𝑥 𝑛+1
𝑛+1
𝑛
+ 𝑐; 𝑛 ≠ −1
∫(𝑎𝑥 + 𝑏) 𝑑𝑥 =
(𝑎𝑥+𝑏)𝑛+1
𝑎.(𝑛+1)
+ 𝑐; 𝑛 ≠ −1
Integrals of a Combination of Functions
Theorem 1: The integral of the product of a constant and a function is equal to the
product of the constant and the integral of the function.
∫ 𝑘. 𝑓 (𝑥 )𝑑𝑥 = 𝑘 ∫ 𝑓 (𝑥 )𝑑𝑥
Theorem 2: The integral of a sum or difference of two functions is equal to the sum
or difference of their integrals.
∫[𝑓(𝑥 ) ± 𝑔(𝑥 )]𝑑𝑥 = ∫ 𝑓 (𝑥 )𝑑𝑥 ± ∫ 𝑔(𝑥 )𝑑𝑥
Integrals of Trigonometric Functions
1.
∫ sin 𝑥 𝑑𝑥 = − cos 𝑥 + 𝑐
2.
∫ cos 𝑥 𝑑𝑥 = sin 𝑥 + 𝑐
3.
∫ sec 2 𝑥 𝑑𝑥 = tan 𝑥 + 𝑐
4.
∫ cosec 2 𝑥 𝑑𝑥 = − cot 𝑥 + 𝑐
5.
∫ sec 𝑥 tan 𝑥 𝑑𝑥 = sec 𝑥 + 𝑐
6.
∫ 𝑐𝑜𝑠𝑒𝑐 𝑥 cot 𝑥 𝑑𝑥 = −𝑐𝑜𝑠𝑒𝑐 𝑥 + 𝑐
7.
∫ tan 𝑥 𝑑𝑥 = log|sec 𝑥 | + 𝑐
8.
∫ cot 𝑥 𝑑𝑥 = log|sin 𝑥 | + 𝑐
9.
∫ sec 𝑥 𝑑𝑥 = log|sec 𝑥 + tan 𝑥 | + 𝑐 = log |tan ( 4 + 2)| + 𝑐
𝜋
𝑥
𝑥
10. ∫ cosec 𝑥 𝑑𝑥 = log|cosec 𝑥 − cot 𝑥 | + 𝑐 = log |tan | + 𝑐
2
Formulae and Identities of Trigonometrical Functions
The following identities and relations can be used to simplify the integrand when it contains higher powers of
sin x, cos x and other t-functions and hence bring it into any of the standard forms:
1.
Fundamental Identities
(a) Square Relations: 1) 𝒔𝒊𝒏𝟐 𝜽 + 𝒄𝒐𝒔𝟐 𝜽 = 𝟏 ⇒ 𝒔𝒊𝒏𝟐 𝜽 = 𝟏 − 𝒄𝒐𝒔𝟐 𝜽 ⇒ 𝒄𝒐𝒔𝟐 𝜽 = 𝟏 − 𝒔𝒊𝒏𝟐 𝜽
2) 𝟏 + 𝒕𝒂𝒏𝟐 𝜽 = 𝒔𝒆𝒄𝟐 𝜽 ⇒ 𝒕𝒂𝒏𝟐 𝜽 = 𝒔𝒆𝒄𝟐 𝜽 − 𝟏 ⇒ 𝒔𝒆𝒄𝟐 𝜽 − 𝒕𝒂𝒏𝟐 𝜽 = 𝟏
3) 𝟏 + 𝒄𝒐𝒕𝟐 𝜽 = 𝒄𝒐𝒔𝒆𝒄𝟐 𝜽 ⇒ 𝒄𝒐𝒕𝟐 𝜽 = 𝒄𝒐𝒔𝒆𝒄𝟐 𝜽 − 𝟏 ⇒ 𝒄𝒐𝒔𝒆𝒄𝟐 𝜽 − 𝒄𝒐𝒕𝟐 𝜽 = 𝟏
(b)Basic Relations: 1)
𝒔𝒊𝒏 𝒙
𝒄𝒐𝒔𝟐 𝒙
𝒕𝒂𝒏𝒙
3)
2.
𝒔𝒆𝒄 𝒙
= 𝒔𝒆𝒄 𝒙 𝒕𝒂𝒏 𝒙
= 𝒔𝒊𝒏 𝒙
4)
2)
𝒄𝒐𝒕𝒙
𝒄𝒐𝒔𝒆𝒄 𝒙
Converting products into sums or differences
Using Corollaries from Multiple Angle Forms
(i) 𝒔𝒊𝒏𝟐 𝑨 =
𝟏−𝒄𝒐𝒔 𝟐𝑨
(ii) 𝒄𝒐𝒔𝟐 𝑨 =
𝟏+𝒄𝒐𝒔 𝟐𝑨
𝟐
𝟐
𝟏−𝒄𝒐𝒔 𝟐𝑨
(iii) 𝟏+𝒄𝒐𝒔 𝟐𝑨 = 𝒕𝒂𝒏𝟐 𝑨 = 𝒔𝒆𝒄𝟐 𝑨 − 𝟏
(iv) √𝟏 ± 𝒔𝒊𝒏 𝟐𝑨 = 𝒄𝒐𝒔 𝑨 ± 𝒔𝒊𝒏 𝑨
(v)
𝟏− 𝒄𝒐𝒔 𝑨
(vi)
𝒔𝒊𝒏 𝑨
𝟏+ 𝒄𝒐𝒔 𝑨
𝒔𝒊𝒏 𝑨
=
𝟐 𝒔𝒊𝒏𝟐
𝑨
𝟐
𝑨
𝟐
𝟐 𝒔𝒊𝒏 𝒄𝒐𝒔
=
𝑨
𝟐
𝟐 𝒄𝒐𝒔𝟐
𝑨
𝟐
𝑨
𝑨
𝟐 𝒔𝒊𝒏 𝒄𝒐𝒔
𝟐
𝟐
𝟏
𝟏
𝑨
𝟐
𝑨
𝒄𝒐𝒔
𝟐
𝒔𝒊𝒏
=
=
= 𝒕𝒂𝒏
𝑨
𝟐
𝑨
𝒔𝒊𝒏
𝟐
𝒄𝒐𝒔
𝑨
(vii) ) 𝟏+𝒄𝒐𝒔 𝟐𝑨 = 𝟐 𝒔𝒆𝒄𝟐 𝟐
𝟏
𝟏
𝑨
(viii) 𝟏−𝒄𝒐𝒔 𝟐𝑨 = 𝟐 𝒄𝒐𝒔𝒆𝒄𝟐 𝟐
𝟏
(ix) 𝒔𝒊𝒏𝟑 𝑨 = 𝟒 [𝟑𝒔𝒊𝒏 𝑨 − 𝒔𝒊𝒏 𝟑𝑨]
𝟏
(x) 𝒄𝒐𝒔𝟑 𝑨 = 𝟒 [𝒄𝒐𝒔 𝟑𝑨 + 𝟑 𝒄𝒐𝒔 𝑨]
𝑨
𝟐
= 𝒄𝒐𝒕
𝑨
𝟐
= 𝒄𝒐𝒔𝒆𝒄 𝒙 𝒄𝒐𝒕 𝒙
= 𝒄𝒐𝒔 𝒙
(i) 𝟐 𝒔𝒊𝒏 𝑨 𝒄𝒐𝒔 𝑩 = 𝒔𝒊𝒏 (𝑨 + 𝑩) + 𝒔𝒊𝒏 (𝑨 − 𝑩)
(ii) 𝟐 𝒄𝒐𝒔 𝑨 𝒔𝒊𝒏 𝑩 = 𝒔𝒊𝒏 (𝑨 + 𝑩) − 𝒔𝒊𝒏 (𝑨 − 𝑩)
(iii) 𝟐 𝒄𝒐𝒔 𝑨 𝒄𝒐𝒔 𝑩 = 𝒄𝒐𝒔 (𝑨 + 𝑩) + 𝒄𝒐𝒔(𝑨 − 𝑩)
(iv) 𝟐 𝒔𝒊𝒏 𝑨 𝒔𝒊𝒏 𝑩 = 𝒄𝒐𝒔 (𝑨 − 𝑩) − 𝒄𝒐𝒔 (𝑨 + 𝑩)
3.
𝒄𝒐𝒔 𝒙
𝒔𝒊𝒏𝟐 𝒙
𝟏
Integrals of ,
𝟏
𝒙 𝒂𝒙+𝒃
, 𝒆𝒙 , 𝒂𝒙 𝒂𝒏𝒅 𝒆𝒂𝒙+𝒃
1
1.
∫ 𝑥 𝑑𝑥 = log|𝑥 | + 𝑐
2.
∫ 𝑎𝑥+𝑏 𝑑𝑥 = a log|𝑎𝑥 + 𝑏| + 𝑐
3.
∫ 𝑒 𝑥 𝑑𝑥 = 𝑒 𝑥 + 𝑐
4.
∫ 𝑎 𝑥 𝑑𝑥 = log 𝑎 + 𝑐
5.
∫ 𝑒 𝑎𝑥+𝑏 𝑑𝑥 =
1
1
𝑎𝑥
𝑒 𝑎𝑥+𝑏
𝑎
+𝑐
Integrals of Inverse Trigonometric Functions
1. ∫
2. ∫
3. ∫
4. ∫
5. ∫
6. ∫
1
√1−𝑥 2
1
1+𝑥 2
𝑑𝑥 = sin−1 𝑥 + 𝑐 = − cos −1 𝑥 + 𝑐
𝑑𝑥 = tan−1 𝑥 + 𝑐 = − cot −1 𝑥 + 𝑐
1
|𝑥|√𝑥 2 −1
1
𝑑𝑥 = sec −1 𝑥 + 𝑐 = − cosec −1 𝑥 + 𝑐
𝑥
√𝑎2 −𝑥
1
𝑎2 +𝑥
𝑑𝑥 = sin−1 + 𝑐
2
𝑎
1
𝑥
a
𝑎
−1
+𝑐
2 𝑑𝑥 = tan
1
|𝑥|√𝑥 2 −𝑎
1
𝑥
a
𝑎
𝑑𝑥 = sec −1 + 𝑐
2
Note: Learn all properties of inverse trigonometric functions very well.
STANDARD METHODS OF INTEGARTION
When the given function cannot be integrated directly by using standard formulae, we
try other methods of integration. The process of integration is largely of tentative
nature and no systematic procedure can be given as in differentiation. However the
following are two important methods of integration:
(i) Integration by substitution
(ii) Integration by parts
Integration by Substitution
Consider the following complicated integrand. We notice that the given integrand
cannot be integrated directly by using any of the known standard forms. So, we use
the method of substitution to change the variable of integration so that the integrand
becomes simple enough to be integrated by applying a known standard formula.
∫[𝑓(𝑥 )]𝑛 𝑓 ′ (𝑥 )𝑑𝑥 ; 𝐿𝑒𝑡 𝑓 (𝑥 ) = 𝑡
𝑓 ′ (𝑥 )𝑑𝑥 = 𝑑𝑡
The integrand now takes on the form,
∫ 𝑡 𝑛 𝑑𝑡 =
𝑡 𝑛+1
𝑛+1
+𝑐 =
[𝑓(𝑥)]𝑛+1
𝑛+1
+𝑐
Note: There are no hard and fast rules for making suitable substitution. Practice will
help acquire the necessary skill. However, the following guidelines will be found
useful:
(i)
If the integrand contains
√1 − 𝑥 2 ; 𝑠𝑢𝑏𝑠𝑡𝑖𝑡𝑢𝑡𝑒 𝑥 = sin 𝜃 𝑜𝑟 𝑥 = cos 𝜃
√𝑎2 − 𝑥 2 ; 𝑠𝑢𝑏𝑠𝑡𝑖𝑡𝑢𝑡𝑒 𝑥 = asin 𝜃 𝑜𝑟 𝑥 = acos 𝜃
(ii)
If the integrand contains
(1 + 𝑥 2 ) 𝑜𝑟 √1 + 𝑥 2 ; 𝑠𝑢𝑏𝑠𝑡𝑖𝑡𝑢𝑡𝑒 𝑥 = tan 𝜃 𝑜𝑟 𝑥 = cot 𝜃
(𝑎2 + 𝑥 2 ) 𝑜𝑟 √𝑎2 + 𝑥 2 ; 𝑠𝑢𝑏𝑠𝑡𝑖𝑡𝑢𝑡𝑒 𝑥 = atan 𝜃 𝑜𝑟 𝑥 = acot 𝜃
(iii) If the integrand contains
√𝑥 2 − 1 ; 𝑠𝑢𝑏𝑠𝑡𝑖𝑡𝑢𝑡𝑒 𝑥 = sec 𝜃 𝑜𝑟 𝑥 = cosec 𝜃
√𝑥 2 − 𝑎2 ; 𝑠𝑢𝑏𝑠𝑡𝑖𝑡𝑢𝑡𝑒 𝑥 = asec 𝜃 𝑜𝑟 𝑥 = acosec 𝜃
(iv)
If the integrand contains
√1 ± 𝑥 ; 𝑠𝑢𝑏𝑠𝑡𝑖𝑡𝑢𝑡𝑒 𝑥 = sin 2𝜃 𝑜𝑟 𝑥 = cos 2𝜃
𝑜𝑟 𝑠𝑢𝑏𝑠𝑡𝑖𝑡𝑢𝑡𝑒 (1 ± 𝑥 ) = 𝑡 2 𝑜𝑟 √1 ± 𝑥 = 𝑡
√𝑎 ± 𝑥 ; 𝑠𝑢𝑏𝑠𝑡𝑖𝑡𝑢𝑡𝑒 𝑥 = asin 2𝜃 𝑜𝑟 𝑥 = acos 2𝜃
𝑜𝑟 𝑠𝑢𝑏𝑠𝑡𝑖𝑡𝑢𝑡𝑒 (𝑎 ± 𝑥 ) = 𝑡 2 𝑜𝑟 √𝑎 ± 𝑥 = 𝑡
(v)
If the integrand contains
√𝑎−𝑥
; 𝑠𝑢𝑏𝑠𝑡𝑖𝑡𝑢𝑡𝑒
√𝑎+𝑥
(vi)
𝑥 = acos 2𝜃
If the integrand contains some linear expression of x, put that value= 𝑡.
(vii) ∫
cos 𝑥−sin 𝑥
cos 𝑥+sin 𝑥
𝑑𝑥 ; 𝑃𝑢𝑡 (cos 𝑥 + sin 𝑥) = 𝑡 ⇒ (cos 𝑥 − sin 𝑥)𝑑𝑥 = 𝑑𝑡
𝑒 𝑥 −𝑒 −𝑥
(viii) ∫ 𝑥 −𝑥 𝑑𝑥 ; 𝑃𝑢𝑡 (𝑒 𝑥 + 𝑒 −𝑥 ) = 𝑡 ⇒ (𝑒 𝑥 − 𝑒 −𝑥 )𝑑𝑥 = 𝑑𝑡
𝑒 +𝑒
(ix) If the integrand contains √𝑥 or similar terms, we can put √𝑥 = 𝑡 so that
𝑑𝑥 = 2𝑡𝑑𝑡
Some Important Formulas
1. ∫[𝑓(𝑥
2. ∫
)]𝑛 ′ (
𝑓 ′ (𝑥)
𝑓(𝑥)
𝑓 𝑥 )𝑑𝑥 (𝑛 ≠ −1) =
[𝑓(𝑥)]𝑛+1
𝑛+1
+𝑐
𝑑𝑥 = log 𝑓 (𝑥 ) + 𝑐
Note: Do not directly use the above two formulae. Make the necessary substitution,
integrate in terms of the new variable and finally re-substitute back the answer in
terms of the starting variable.
Integration by Parts
If 𝒖 and 𝒗 are two functions of 𝒙, then
𝒅𝒖
∫ 𝒖𝒗 𝒅𝒙 = 𝒖 ∫ 𝒗 𝒅𝒙 − ∫ [ 𝒅𝒙 . ∫ 𝒗 𝒅𝒙] 𝒅𝒙
i.e.
𝑰𝒏𝒕𝒆𝒈𝒓𝒂𝒍 𝒐𝒇 𝒕𝒉𝒆 𝒑𝒓𝒐𝒅𝒖𝒄𝒕 𝒐𝒇 𝒕𝒘𝒐 𝒇𝒖𝒄𝒏𝒕𝒊𝒐𝒏𝒔 = 𝟏𝒔𝒕 𝒇𝒖𝒄𝒏𝒕𝒊𝒐𝒏 ×
𝑰𝒏𝒕𝒆𝒈𝒓𝒂𝒍 𝒐𝒇 𝟐𝒏𝒅 − 𝑰𝒏𝒕𝒆𝒈𝒓𝒂𝒍 𝒐𝒇 [𝑫𝒊𝒇𝒇𝒆𝒓𝒆𝒏𝒕𝒊𝒂𝒍 𝒐𝒇 𝟏𝒔𝒕 × 𝑰𝒏𝒕𝒆𝒈𝒓𝒂𝒍 𝒐𝒇 𝟐𝒏𝒅 ]
If the two functions are of different types, take the first function to be the function
which comes first in the word ‘ILATE’ where
I – Inverse Trigonometric Functions
L – Logarithmic Functions
A – Algebraic Functions
T – Trigonometric Functions
E – Exponential Functions
Note:
 The rule of integration by parts is useful in integrating single functions like
logarithmic and inverse t-functions. For this purpose, unity (i.e. 1) is taken as the
second function.
 Here also, we see that ILATE Rule is not completely foolproof or hard and fast.
However it will found to be useful in most cases. Sometimes, not following ILATE
Rule is found to be simpler.
 In many cases, substitution precedes the use of the by-parts formula.
Integrals of the form ∫ 𝒆𝒂𝒙 𝒔𝒊𝒏 (𝒃𝒙 + 𝒄) 𝒅𝒙 or ∫ 𝒆𝒂𝒙 𝒄𝒐𝒔 (𝒃𝒙 + 𝒄) 𝒅𝒙
In such cases, we first put 𝐼 = ∫ 𝑒 𝑎𝑥 𝑠𝑖𝑛 (𝑏𝑥 + 𝑐) 𝑑𝑥 or 𝐼 = ∫ 𝑒 𝑎𝑥 𝑐𝑜𝑠 (𝑏𝑥 + 𝑐) 𝑑𝑥.
Then we integrate the given function by parts taking 𝒆𝒂𝒙 as first function and
𝒔𝒊𝒏 (𝒃𝒙 + 𝒄) or 𝒄𝒐𝒔 (𝒃𝒙 + 𝒄) as the second function. We then notice that on
integrating by parts the second time, the integrand contains a term of 𝐼. By taking all
the 𝐼 terms to one side, we get the value of 𝐼, which is the necessary integral.
Integrals of the form ∫ 𝒆𝒙 [𝒇(𝒙) + 𝒇′ (𝒙)] 𝒅𝒙
Method:
1. First split the given integral into two parts as follows:
∫ 𝑒 𝑥 [𝑓(𝑥 ) + 𝑓 ′ (𝑥 )] 𝑑𝑥 = ∫ 𝑒 𝑥 𝑓(𝑥 ) 𝑑𝑥 + ∫ 𝑒 𝑥 𝑓 ′ (𝑥 ) 𝑑𝑥
2. Integrate only the first integral i.e. ∫ 𝑒 𝑥 𝑓(𝑥 ) 𝑑𝑥 by parts taking 𝑓(𝑥 ) as first
function and 𝑒 𝑥 as second function.
3. Now, the integrand becomes 𝑓(𝑥 )𝑒 𝑥 − ∫ 𝑒 𝑥 𝑓 ′ (𝑥 ) 𝑑𝑥 + ∫ 𝑒 𝑥 𝑓 ′ (𝑥 ) 𝑑𝑥 . The
last two integrals will cancel each other.
Integrating a Complex Fraction by Decomposing into Partial Fractions
There are various methods of decomposing a given proper rational fraction into a
group of simple fractions each having as its denominator one of the factors of the
denominator of the original rational fraction. These simple fractions are called partial
fractions. Partial fractions method can be used in the following cases:
Case 1: When the denominator of the original complex fraction contains linear
unrepeated factors.
𝑥+𝑎
𝐴
𝐵
=
+
(𝑥 + 𝑏)(𝑥 + 𝑐) (𝑥 + 𝑏) (𝑥 + 𝑐)
We then solve the given expression by substituting various values of 𝑥 to obtain the
values of A and B.
This method can be extended and the given function can be decomposed to more
number of partial fractions depending upon the number of linear factors present in
the denominator of the given function.
Case 2: When the denominator of the original complex fraction contains linear
repeated factors.
𝑥+𝑎
𝐴
𝐵
𝐶
=
+
+
(𝑥 + 𝑏)(𝑥 + 𝑐)2 (𝑥 + 𝑏) (𝑥 + 𝑐) (𝑥 + 𝑐)2
We then solve the given expression by substituting various values of 𝑥 to obtain the
values of A, B and C.
This method can be extended and the given function can be decomposed to more
number of partial fractions depending upon the number of linear factors present in
the denominator of the given function.
Case 3: When the denominator of the original complex fraction contains a quadratic
equation not resolvable into linear factors.
𝑥+𝑙
𝐴
𝐵𝑥 + 𝐶
=
+
(𝑚𝑥 + 𝑛)(𝑎𝑥 2 + 𝑏𝑥 + 𝑐 ) (𝑚𝑥 + 𝑛) (𝑎𝑥 2 + 𝑏𝑥 + 𝑐)
We then solve the given expression by substituting various values of 𝑥 to obtain the
values of A, B and C. We can also equate the corresponding coefficients on both
sides to find the above values.
This method can be extended and the given function can be decomposed to more
number of partial fractions depending upon the number of linear or quadratic factors
present in the denominator of the given function.
Case 4: When there are polynomials in the numerator and denominator and the
degree of the numerator is greater than the degree of the denominator i.e. the given
fraction is an improper fraction, then we first use polynomial division to convert it
into a proper fraction and then apply any of the above cases if needed.
INTEGRALS OF THE FORM ∫
𝑑𝑥
1
𝑥
𝑑𝑥
1
𝑥−𝑎
𝑑𝑥
1
𝑎+𝑥
𝒅𝒙
𝒙𝟐 +𝒂
𝟐,∫
𝒅𝒙
𝒙𝟐 −𝒂
𝟐,∫
𝒅𝒙
𝒂𝟐 −𝒙𝟐
AND APPLICATIONS
1. ∫ 2 2 = tan−1 + 𝑐
𝑥 +𝑎
𝑎
𝑎
2. ∫ 2 2 = 𝑙𝑜𝑔 | | + 𝑐
𝑥 −𝑎
2𝑎
𝑥+𝑎
3. ∫ 2 2 = 𝑙𝑜𝑔 | | + 𝑐
𝑎 −𝑥
2𝑎
𝑎−𝑥
Note: Before applying the above standard forms, always make the coefficient of 𝑥 2
unity.
Method for solving integrals of the form ∫
𝒙𝟐 ±𝟏
𝒙𝟒 +𝟏
𝒅𝒙:
Working Rule:
1. Divide the numerator and denominator by 𝑥 2 .
1
2. It now takes the form ∫
1± 2
𝑥
1
𝑥 2+ 2
𝑥
1
1
𝑥
𝑥2
𝑑𝑥. Now substitute 𝑥 ∓ = 𝑡 ⇒ (1 ±
2
3. To get an expression for (𝑥 +
1
𝑥
1 2
1
𝑥
𝑥2
2
2
2 ), (𝑥 ∓ ) = 𝑡 ⇒ (𝑥 +
4. Rewrite the given integrand in terms of t and integrate using ∫
) 𝑑𝑥 = 𝑑𝑡.
) = 𝑡 2 ± 2.
𝑑𝑥
𝑥 2 +𝑎2
,∫
𝑑𝑥
𝑥 2 −𝑎2
form.
Note: The method used above can be generalized to integrate integrals of the form
𝑥 2 ±𝑎
∫ 𝑥 4 +𝑘𝑥 2 +𝑐 𝑑𝑥.
Method of Completing Squares
This method can be used to integrate integrals consisting of quadratic equations in
the denominator such as ∫
𝑑𝑥
𝑞𝑢𝑎𝑑𝑟𝑎𝑡𝑖𝑐
i.e. ∫
𝑑𝑥
𝑎𝑥 2 +𝑏𝑥+𝑐
etc.
This method is especially useful in integrating quadratic equations which are not
resolvable into linear factors.
Working Rule:
1. Make the coefficient of 𝑥 2 unity by taking the numerical coefficient of 𝑥 2 outside.
𝑑𝑥
1
∫ 𝑎𝑥 2 +𝑏𝑥+𝑐 = 𝑎 ∫
𝑑𝑥
𝑏
𝑎
𝑥 2 + 𝑥+
𝑐
𝑎
.
2. Complete squares in terms containing 𝑥 by adding and subtracting the square of
half of the coefficient of 𝑥 and put the denominator in the form {(𝑥 + 𝛼 )2 ± 𝛽 2 }
1
∫
𝑎
1
𝑑𝑥
𝑏
𝑐
𝑥 2 +𝑎𝑥+𝑎
1
= ∫
𝑎
𝑑𝑥
𝑏
𝑏2 𝑏2 𝑐
𝑥 2 + 𝑥+ − +
𝑎
4𝑎 4𝑎 𝑎
1
𝑑𝑥
= ∫
𝑎
𝑏 2
𝑏2
(𝑥+2𝑎) +(𝑐−4𝑎)
which is of the form
𝑑𝑥
.
∫
𝑎 (𝑥+𝛼)2 ±𝛽2
3. Apply the appropriate standard form ∫
𝑑𝑥
𝑥 2 +𝑎
Method to evaluate integrals of the form ∫
2,∫
𝑑𝑥
𝑥 2 −𝑎
𝒑𝒙+𝒒
𝒂𝒙𝟐 +𝒃𝒙+𝒄
2,∫
𝑑𝑥
𝑎2 −𝑥 2
𝒅𝒙 i.e. ∫
.
𝒍𝒊𝒏𝒆𝒂𝒓
𝒒𝒖𝒂𝒅𝒓𝒂𝒕𝒊𝒄
𝒅𝒙
Working Rule:
1. Split the given integral into two parts as follows:
𝑝𝑥+𝑞
𝑥
1
∫ 𝑎𝑥 2 +𝑏𝑥+𝑐 𝑑𝑥 = 𝑝 ∫ 𝑎𝑥 2 +𝑏𝑥+𝑐 𝑑𝑥 + 𝑞 ∫ 𝑎𝑥 2 +𝑏𝑥+𝑐 𝑑𝑥 .
2. The first integral can be integrated by the method of substitution by putting 𝑎𝑥 2 +
𝑏𝑥 + 𝑐 = 𝑡 while the second integral can be integrated directly by applying the
method of completing squares and applying the appropriate standard form.
INTEGRALS OF THE FORM ∫
𝑑𝑥
𝒅𝒙
√𝒙𝟐 +𝒂
,∫
𝟐
𝒅𝒙
√𝒙𝟐 −𝒂
,∫
𝟐
𝒅𝒙
& APPLICATIONS
√𝒂𝟐 −𝒙𝟐
𝑥
1. ∫ 2 2 = 𝑠𝑖𝑛−1 + 𝑐
𝑎
√𝑎 −𝑥
𝑑𝑥
2. ∫ 2 2 = log|𝑥 + √𝑥 2 + 𝑎2 | + 𝑐
√𝑥 +𝑎
𝑑𝑥
3. ∫ 2 2 = log|𝑥 + √𝑥 2 − 𝑎2 | + 𝑐
√𝑥 −𝑎
Note: Before applying the above standard forms, always make the coefficient of 𝑥 2
unity.
Method to evaluate integrals of the type ∫
𝒅𝒙
√𝒂𝒙𝟐 +𝒃𝒙+𝒄
i.e ∫
𝒅𝒙
√𝒒𝒖𝒂𝒅𝒓𝒂𝒕𝒊𝒄
Working Rule:
1. Make the coefficient of 𝑥 2 unity by taking the numerical coefficient of 𝑥 2 outside.
𝑑𝑥
∫ √𝑎𝑥 2 +𝑏𝑥+𝑐 =
1
∫
𝑎
√
𝑏
𝑐
𝑎
𝑎
𝑑𝑥
𝑏
𝑎
√𝑥 2 + 𝑥+
𝑐
𝑎
.
2. Put 𝑥 2 + 𝑥 + in the form {(𝑥 + 𝛼 )2 ± 𝛽 2 } by method of completing squares.
3. Apply the appropriate standard form ∫
Method to evaluate integrals of the type ∫
𝑑𝑥
√𝑥 2 +𝑎
,
2 ∫
𝒑𝒙+𝒒
√𝒂𝒙𝟐 +𝒃𝒙+𝒄
𝑑𝑥
√𝑥 2 −𝑎
,
2 ∫
𝒅𝒙 i.e. ∫
𝑑𝑥
√𝑎2 −𝑥 2
.
𝒍𝒊𝒏𝒆𝒂𝒓
√𝒒𝒖𝒂𝒅𝒓𝒂𝒕𝒊𝒄
𝒅𝒙
Working Rule:
1. Put 𝑝𝑥 + 𝑞 = 𝐴.
𝑑
𝑑𝑥
(𝑎𝑥 2 + 𝑏𝑥 + 𝑐) + 𝐵 i.e. 𝑙𝑖𝑛𝑒𝑎𝑟 = 𝐴.
𝑑
𝑑𝑥
(𝑞𝑢𝑎𝑑𝑟𝑎𝑡𝑖𝑐 ) + 𝐵.
2. Find A and B by equating the coefficients of the corresponding terms on both
sides.
3. Rewrite the given integral by substituting the numerator,
𝑝𝑥 + 𝑞 = 𝐴.
𝑑
𝑑𝑥
(𝑎𝑥 2 + 𝑏𝑥 + 𝑐) + 𝐵.
4. Now the given integral takes the form ∫
𝐴.
𝑑
(𝑎𝑥 2 +𝑏𝑥+𝑐)+𝐵
𝑑𝑥
√𝑎𝑥 2 +𝑏𝑥+𝑐
𝑑𝑥.
5. Now, split the above integral into two integrals which can be easily evaluated:
𝐴∫
𝑑
(𝑎𝑥 2 +𝑏𝑥+𝑐)
𝑑𝑥
√𝑎𝑥 2 +𝑏𝑥+𝑐
𝑑𝑥 + 𝐵 ∫
𝑑𝑥
√𝑎𝑥 2 +𝑏𝑥+𝑐
.
6. The first integral can be integrated by the method of substitution by putting 𝑎𝑥 2 +
𝑏𝑥 + 𝑐 = 𝑡 while the second integral can be integrated directly by applying the
method of completing squares and applying the appropriate standard form.
INTEGRALS OF THE FORM ∫ √𝒂𝟐 − 𝒙𝟐 𝒅𝒙, ∫ √𝒙𝟐 + 𝒂𝟐 𝒅𝒙, ∫ √𝒙𝟐 − 𝒂𝟐 𝒅𝒙 &
APPLICATIONS
1
1
1
1
1
1
𝑥
1. ∫ √𝑎2 − 𝑥 2 𝑑𝑥 = 𝑥√𝑎2 − 𝑥 2 + 𝑎2 𝑠𝑖𝑛−1 + 𝑐
2
2
𝑎
2. ∫ √𝑥 2 + 𝑎2 𝑑𝑥 = 𝑥√𝑥 2 + 𝑎2 + a2 log|𝑥 + √𝑥 2 + 𝑎2 | + 𝑐
2
2
3. ∫ √𝑥 2 − 𝑎2 𝑑𝑥 = 𝑥√𝑥 2 + 𝑎2 − a2 log|𝑥 + √𝑥 2 − 𝑎2 | + 𝑐
2
2
Aid to Memory:
1
1
∫(𝐼𝑛𝑡𝑒𝑔𝑟𝑎𝑛𝑑) 𝑑𝑥 = 2 𝑥(𝐼𝑛𝑡𝑒𝑔𝑟𝑎𝑛𝑑) ± 2 a2 (𝐼𝑛𝑡𝑒𝑔𝑟𝑎𝑙 𝑜𝑓 𝑅𝑒𝑐𝑖𝑝𝑟𝑜𝑐𝑎𝑙 𝑜𝑓 𝐼𝑛𝑡𝑒𝑔𝑟𝑎𝑛𝑑) + 𝑐.
Method to find integrals of the type ∫ √𝒂𝒙𝟐 + 𝒃𝒙 + 𝒄 𝒅𝒙 i.e. ∫ √𝒒𝒖𝒂𝒅𝒓𝒂𝒕𝒊𝒄 𝒅𝒙
Working Rule:
1. Make the coefficient of 𝑥 2 unity by taking the numerical coefficient of 𝑥 2 outside.
𝑏
𝑐
∫ √𝑎𝑥 2 + 𝑏𝑥 + 𝑐 𝑑𝑥 = √𝑎 ∫ √𝑥 2 + 𝑎 𝑥 + 𝑎 𝑑𝑥 .
𝑏
𝑐
𝑎
𝑎
2. Put 𝑥 2 + 𝑥 + in the form {(𝑥 + 𝛼 )2 ± 𝛽 2 } by method of completing squares.
3. Apply the appropriate standard form ∫ √𝑎2 − 𝑥 2 𝑑𝑥, ∫ √𝑥 2 + 𝑎2 𝑑𝑥, ∫ √𝑥 2 − 𝑎2 𝑑𝑥.
INTEGRALS OF THE FORM:
𝒅𝒙
𝒅𝒙
𝒅𝒙
𝒅𝒙
∫ 𝒂+𝒃 𝐜𝐨𝐬 𝒙 , ∫ 𝒂+𝒃 𝐬𝐢𝐧 𝒙 , ∫ 𝒂 𝐜𝐨𝐬 𝒙+𝒃 𝐬𝐢𝐧 𝒙 , ∫ 𝒂 𝐜𝐨𝐬 𝒙+𝒃 𝐬𝐢𝐧 𝒙+𝒄 𝒆𝒕𝒄.
Working Rule:
𝑥
1. Put sin 𝑥 =
2 tan2
𝑥
𝑥
1+tan2 2
2𝑥
2. Transfer 1 + tan
2
𝑥
2
Replace 1 + tan
2
, cos 𝑥 =
1−tan2 2
.
𝑥
1+tan2 2
to the numerator after taking L.C.M. of the denominator.
𝑥
= sec 2 in the numerator.
2
𝑥
1
𝑥
2
2
2
3. Put tan = 𝑡 so that sec 2 𝑑𝑥 = 𝑑𝑡.
4. The new integrand will now be in the form ∫
𝑑𝑡
𝑎𝑡 2 +𝑏𝑡+𝑐
which can be evaluated by
the method of completing squares followed by applying the appropriate standard
forms.
INTEGRALS OF THE FORM: ∫
𝒂 𝒄𝒐𝒔 𝒙+𝒃 𝒔𝒊𝒏 𝒙
𝒄 𝐜𝐨𝐬 𝒙+𝒅 𝐬𝐢𝐧 𝒙
𝒅𝒙
Working Rule:
1. Put 𝑎 𝑐𝑜𝑠 𝑥 + 𝑏 𝑠𝑖𝑛 𝑥 = 𝐴.
𝑛𝑢𝑚𝑒𝑟𝑎𝑡𝑜𝑟 = 𝐴.
𝑑
𝑑𝑥
𝑑
𝑑𝑥
(𝑐 cos 𝑥 + 𝑑 sin 𝑥 ) + 𝐵. (𝑐 cos 𝑥 + 𝑑 sin 𝑥 ) i.e.
(𝑑𝑒𝑛𝑜𝑚𝑖𝑛𝑎𝑡𝑜𝑟 ) + 𝐵. (𝑑𝑒𝑛𝑜𝑚𝑖𝑛𝑎𝑡𝑜𝑟 ).
2. Find A and B by equating the coefficients of cos 𝑥 and sin 𝑥 separately on both
sides.
3. Rewrite the given integral by substituting the numerator,
(𝑎 𝑐𝑜𝑠 𝑥 + 𝑏 𝑠𝑖𝑛 𝑥 ) = 𝐴.
𝑑
𝑑𝑥
(𝑐 cos 𝑥 + 𝑑 sin 𝑥 ) + 𝐵(𝑐 cos 𝑥 + 𝑑 sin 𝑥 ).
𝑑
4. Now the given integral takes the form ∫
𝐴.𝑑𝑥(𝑐 cos 𝑥+𝑑 sin 𝑥)+𝐵(𝑐 cos 𝑥+𝑑 sin 𝑥)
𝑐 cos 𝑥+𝑑 sin 𝑥
𝑑𝑥.
5. Now, split the above integral into two integrals as follows:
𝐴∫
𝑑
(𝑐 cos 𝑥+𝑑 sin 𝑥)
𝑑𝑥
𝑐 cos 𝑥+𝑑 sin 𝑥
𝑑𝑥 + 𝐵 ∫ 𝑑𝑥 .
6. The first integral can be integrated by the method of substitution by putting
(𝑐 cos 𝑥 + 𝑑 sin 𝑥) = 𝑡 while the second integral can be integrated directly
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