The document discusses integration and the definition of the definite integral. It provides 30 rules of integration involving trigonometric, logarithmic, exponential and other common functions. It also briefly discusses integration by substitution and defines the process of making a u-substitution to evaluate integrals that can be written in a particular form.

1. Integration & Application
of integration
By GP - Mumbai

2. Contents:
1. Definition of integration as antiderivative
2. Rules of integration
3. Integration by substitution
4. Integration of composite function
5. Definition of definite integral
6. Properties of definite integral with simple problems
7. Area under the curve
8. Area bounded by two curves

3. Definition :
If
𝒅
𝒅𝒙
𝒇 𝒙 + 𝒄 =F(x) then 𝑭 𝒙 𝒅𝒙 = 𝒇 𝒙 + 𝒄 , where c is constant of integration
As
𝒅
𝒅𝒙
𝐜 = 𝟎
and 𝑭(𝒙)dx indicates integration of F(x) with respect to x
Symbol :
𝒅𝒙 𝒊𝒔 𝒐𝒑𝒆𝒓𝒂𝒕𝒐𝒓 𝒊𝒏𝒅𝒊𝒄𝒂𝒕𝒊𝒐𝒏 𝒊𝒏𝒕𝒆𝒈𝒓𝒂𝒕𝒊𝒐𝒏 𝒘𝒊𝒕𝒉 𝒓𝒆𝒔𝒑𝒆𝒄𝒕 𝒕𝒐 𝒙
Example :
𝒅
𝒅𝒙
𝒔𝒊𝒏𝒙 + 𝒄 =𝒄𝒐𝒔𝒙
hence, 𝒄𝒐𝒔𝒙 𝒅𝒙 = 𝒔𝒊𝒏𝒙 + 𝒄
1. Definition of integration as antiderivative

4. 1.
𝒅
𝒅𝒙
(𝒙 𝒏+𝟏
+𝒄) = 𝒏 + 𝟏 𝒙 𝒏
=> 𝒙 𝒏
𝒅𝒙 =
𝒙 𝒏+𝟏
𝒏+𝟏
+ 𝒄 𝒊𝒇 𝒏 ≠ −𝟏
2.
𝒅
𝒅𝒙
𝐥𝐨𝐠 𝒙 + 𝒄 =
𝟏
𝒙
=>
𝟏
𝒙
𝒅𝒙 = 𝐥𝐨𝐠 𝒙 + 𝒄
3.
𝒅
𝒅𝒙
𝒄𝒐𝒔𝒙 + 𝒄 = −𝒔𝒊𝒏𝒙 => 𝒔𝒊𝒏𝒙 𝒅𝒙 = −𝒄𝒐𝒔𝒙 + 𝒄
4.
𝒅
𝒅𝒙
𝒔𝒊𝒏𝒙 + 𝒄 = 𝒄𝒐𝒔𝒙 => 𝒄𝒐𝒔𝒙 𝒅𝒙 = 𝒔𝒊𝒏𝒙 + 𝒄
5.
𝒅
𝒅𝒙
𝐥𝐨𝐠 𝒔𝒆𝒄𝒙 + 𝒄 = 𝒕𝒂𝒏𝒙 => 𝒕𝒂𝒏𝒙 𝒅𝒙 = 𝐥𝐨𝐠 𝒔𝒆𝒄𝒙 + 𝒄 = − 𝐥𝐨𝐠 𝒄𝒐𝒔𝒙 + 𝒄
6.
𝒅
𝒅𝒙
𝐥𝐨𝐠 𝒔𝒊𝒏𝒙 + 𝒄 = 𝒄𝒐𝒕𝒙 => 𝒄𝒐𝒕𝒙 𝒅𝒙 = 𝐥𝐨𝐠 𝒔𝒊𝒏𝒙 + 𝒄
7.
𝒅
𝒅𝒙
𝒕𝒂𝒏𝒙 + 𝒄 = 𝒔𝒆𝒄 𝟐
𝒙 => 𝒔𝒆𝒄 𝟐
𝒙 𝒅𝒙 = 𝒕𝒂𝒏𝒙 + 𝒄
8.
𝒅
𝒅𝒙
−𝒄𝒐𝒕𝒙 + 𝒄 = 𝒄𝒐𝒔𝒆𝒄 𝟐
𝒙 => 𝒄𝒐𝒔𝒆𝒄 𝟐
𝒙 𝒅𝒙 = −𝒄𝒐𝒕𝒙 + 𝒄
2. Rules of integration Part -1

5. 9.
𝒅
𝒅𝒙
𝒔𝒆𝒄𝒙 + 𝒄 = 𝒔𝒆𝒄𝒙. 𝒕𝒂𝒏𝒙 => 𝒔𝒆𝒄𝒙. 𝒕𝒂𝒏𝒙 𝒅𝒙 = 𝒔𝒆𝒄𝒙 + 𝒄
10.
𝒅
𝒅𝒙
−𝒄𝒐𝒔𝒆𝒄𝒙 + 𝒄 = 𝒄𝒐𝒔𝒆𝒄𝒙. 𝒄𝒐𝒕𝒙 => 𝒄𝒐𝒔𝒆𝒄𝒙. 𝒄𝒐𝒕𝒙 𝒅𝒙 = −𝒄𝒐𝒔𝒆𝒄𝒙 + 𝒄
11.
𝒅
𝒅𝒙
𝒍𝒐𝒈 𝒄𝒐𝒔𝒆𝒄𝒙 − 𝒄𝒐𝒕𝒙 + 𝒄 = 𝒄𝒐𝒔𝒆𝒄𝒙 => 𝒄𝒐𝒔𝒆𝒄𝒙 𝒅𝒙 = 𝒍𝒐𝒈 𝒄𝒐𝒔𝒆𝒄𝒙 − 𝒄𝒐𝒕𝒙 + 𝒄
12.
𝒅
𝒅𝒙
𝒍𝒐𝒈 𝒔𝒆𝒄𝒙 + 𝒕𝒂𝒏𝒙 + 𝒄 = 𝒔𝒆𝒄𝒙 => 𝒔𝒆𝒄𝒙 𝒅𝒙 = 𝒍𝒐𝒈 𝒔𝒆𝒄𝒙 + 𝒕𝒂𝒏𝒙 + 𝒄
13.
𝒅
𝒅𝒙
𝒆 𝒙 + 𝒄 = 𝒆 𝒙 => 𝒆 𝒙 𝒅𝒙 = 𝒆 𝒙 + 𝒄
14.
𝒅
𝒅𝒙
𝒂 𝒙
𝐥𝐨𝐠 𝒆 𝒂
+ 𝒄 = 𝒂 𝒙
=> 𝒂 𝒙
𝒅𝒙 =
𝒂 𝒙
𝐥𝐨𝐠 𝒆 𝒂
+ 𝒄
15.
𝒅
𝒅𝒙
𝒔𝒊𝒏−𝟏 𝒙
𝒂
+ 𝒄 =
𝟏
𝒂 𝟐−𝒙 𝟐
=>
𝟏
𝒂 𝟐−𝒙 𝟐
𝒅𝒙 = 𝒔𝒊𝒏−𝟏 𝒙
𝒂
+ 𝒄
2. Rules of integration Part -2

6. 16.
𝒅
𝒅𝒙
𝒄𝒐𝒔−𝟏 𝒙
𝒂
+ 𝒄 =
−𝟏
𝒂 𝟐−𝒙 𝟐
=>
−𝟏
𝒂 𝟐−𝒙 𝟐
𝒅𝒙 = 𝒄𝒐𝒔−𝟏 𝒙
𝒂
+ 𝒄
17.
𝒅
𝒅𝒙
𝟏
𝒂
𝒕𝒂𝒏−𝟏 𝒙
𝒂
+ 𝒄 =
𝟏
𝒂 𝟐+𝒙 𝟐 =>
𝟏
𝒂 𝟐+𝒙 𝟐 𝒅𝒙 =
𝟏
𝒂
𝒕𝒂𝒏−𝟏 𝒙
𝒂
+ 𝒄
18.
𝒅
𝒅𝒙
𝟏
𝒂
𝒄𝒐𝒕−𝟏 𝒙
𝒂
+ 𝒄 =
−𝟏
𝒂 𝟐+𝒙 𝟐 =>
−𝟏
𝒂 𝟐+𝒙 𝟐 𝒅𝒙 =
𝟏
𝒂
𝒄𝒐𝒕−𝟏 𝒙
𝒂
+ 𝒄
19.
𝒅
𝒅𝒙
𝟏
𝒂
𝒔𝒆𝒄−𝟏 𝒙
𝒂
+ 𝒄 =
𝟏
𝒙 𝒙 𝟐−𝒂 𝟐
=>
𝟏
𝒙 𝒙 𝟐−𝒂 𝟐
𝒅𝒙 =
𝟏
𝒂
𝒔𝒆𝒄−𝟏 𝒙
𝒂
+ 𝒄
20.
𝒅
𝒅𝒙
𝟏
𝒂
𝒄𝒐𝒔𝒆𝒄−𝟏 𝒙
𝒂
+ 𝒄 =
−𝟏
𝒙 𝒙 𝟐−𝒂 𝟐
=>
−𝟏
𝒙 𝒙 𝟐−𝒂 𝟐
𝒅𝒙 =
𝟏
𝒂
𝒄𝒐𝒔𝒆𝒄−𝟏 𝒙
𝒂
+ 𝒄
21.
𝟏
𝒂 𝟐−𝒙 𝟐 𝒅𝒙 =
𝟏
𝟐𝒂
𝐥𝐨𝐠
𝒂+𝒙
𝒂−𝒙
+ 𝒄
22.
𝟏
𝒙 𝟐−𝒂 𝟐 𝒅𝒙 =
𝟏
𝟐𝒂
𝐥𝐨𝐠
𝒙−𝒂
𝒙+𝒂
+ 𝒄
23.
𝟏
𝒙 𝟐+𝒂 𝟐
𝒅𝒙 = 𝐥𝐨𝐠 𝒙 + 𝒙 𝟐 + 𝒂 𝟐 + 𝒄
2. Rules of integration Part -3

7. 24.
𝟏
𝒙 𝟐−𝒂 𝟐
𝒅𝒙 = 𝐥𝐨𝐠 𝒙 + 𝒙 𝟐 − 𝒂 𝟐 + 𝒄
25. 𝒂 𝟐 − 𝒙 𝟐 𝒅𝒙 =
𝒙
𝟐
𝒂 𝟐 − 𝒙 𝟐 +
𝒂 𝟐
𝟐
𝒔𝒊𝒏−𝟏 𝒙
𝒂
+ 𝒄
26. 𝒙 𝟐 + 𝒂 𝟐 𝒅𝒙 =
𝒙
𝟐
𝒙 𝟐 + 𝒂 𝟐 +
𝒂 𝟐
𝟐
𝐥𝐨𝐠 𝒙 + 𝒙 𝟐 + 𝒂 𝟐 + 𝒄
27. 𝒙 𝟐 − 𝒂 𝟐 𝒅𝒙 =
𝒙
𝟐
𝒙 𝟐 − 𝒂 𝟐 −
𝒂 𝟐
𝟐
𝐥𝐨𝐠 𝒙 + 𝒙 𝟐 − 𝒂 𝟐 + 𝒄
28. 𝒇 𝒙 𝒏
𝒇 𝒙 𝒅𝒙 =
𝒇(𝒙) 𝒏+𝟏
𝒏+𝟏
+ 𝒄 𝒊𝒇 𝒏 ≠ −𝟏
29.
𝒇′(𝒙)
𝒇(𝒙)
𝒅𝒙 = 𝐥𝐨𝐠 𝒇(𝒙) + 𝒄
30. 𝑰𝒇 𝒇 𝒙 𝒅𝒙 = 𝑭 𝒙 + 𝒄 𝒕𝒉𝒆𝒏 𝒇 𝒂𝒙 + 𝒃 =
𝟏
𝒂
𝑭 𝒂𝒙 + 𝒃 + 𝒄
2. Rules of integration Part -4

8. The first and most vital step is to be able to write your integral in this form:
Note This Step:
3. Integration by substitution
For example :
Here 𝑓 = 𝑐𝑜𝑠 and you have 𝑔 = 𝑥2
and its derivative of 2x
Now integrate:
∫cos(u) du = sin(u) + C
And finally put u= 𝒙 𝟐
back again:
sin(𝑥2) + C

9. 𝟏. 𝒂𝒙 + 𝒃 𝒏 𝒅𝒙 =
𝟏
𝒂
𝒂𝒙+𝒃 𝒏+𝟏
𝒏+𝟏
+ 𝒄 𝒊𝒇 𝒏 ≠ −𝟏
2.
𝟏
𝒂𝒙+𝒃
𝒅𝒙 =
𝟏
𝒂
𝐥𝐨𝐠 𝒂𝒙 + 𝒃 + 𝒄
𝟑. 𝒔𝒊𝒏 𝒂𝒙 + 𝒃 𝒅𝒙 = −
𝟏
𝒂
𝐜𝐨𝐬(𝒂𝒙 + 𝒃) + 𝒄
4. 𝐜𝐨𝐬(𝒂𝒙 + 𝒃) 𝒅𝒙 =
𝟏
𝒂
𝐬𝐢𝐧(𝒂𝒙 + 𝒃) + 𝒄
5. 𝐭𝐚𝐧(𝒂𝒙 + 𝒃) 𝒅𝒙 =
𝟏
𝒂
𝐥𝐨𝐠 𝐬𝐞𝐜(𝒂𝒙 + 𝒃) + 𝒄 = −
𝟏
𝒂
𝐥𝐨𝐠 𝒄𝒐𝒔 𝒂𝒙 + 𝒃 + 𝒄
6. 𝐜𝐨𝐭(𝒂𝒙 + 𝒃) 𝒅𝒙 =
𝟏
𝒂
𝐥𝐨𝐠 𝐬𝐢𝐧(𝒂𝒙 + 𝒃) + 𝒄
7. 𝒔𝒆𝒄 𝟐(𝒂𝒙 + 𝒃) 𝒅𝒙 =
𝟏
𝒂
𝐭𝐚𝐧(𝒂𝒙 + 𝒃) + 𝒄
4. Integration of composite function Part -1

10. 8. 𝒄𝒐𝒔𝒆𝒄 𝟐(𝒂𝒙 + 𝒃) 𝒅𝒙 = −
𝟏
𝒂
𝐜𝐨𝐭(𝒂𝒙 + 𝒃) + 𝒄
9. 𝐬𝐞𝐜(𝒂𝒙 + 𝒃). 𝐭𝐚𝐧(𝒂𝒙 + 𝒃) 𝒅𝒙 =
𝟏
𝒂
𝐬𝐞𝐜(𝒂𝒙 + 𝒃) + 𝒄
10. 𝒄𝒐𝒔𝒆𝒄(𝒂𝒙 + 𝒃). 𝐜𝐨𝐭(𝒂𝒙 + 𝒃) 𝒅𝒙 = −
𝟏
𝒂
𝒄𝒐𝒔𝒆𝒄(𝒂𝒙 + 𝒃) + 𝒄
11. 𝒄𝒐𝒔𝒆𝒄(𝒂𝒙 + 𝒃) 𝒅𝒙 =
𝟏
𝒂
𝐥𝐨𝐠 𝒄𝒐𝒔𝒆𝒄 𝒂𝒙 + 𝒃 − 𝒄𝒐𝒕 𝒂𝒙 + 𝒃 + 𝒄
12. 𝐬𝐞𝐜(𝒂𝒙 + 𝒃) 𝒅𝒙 =
𝟏
𝒂
𝐥𝐨𝐠 𝐬𝐞𝐜(𝒂𝒙 + 𝒃) + 𝐭𝐚𝐧(𝒂𝒙 + 𝒃) + 𝒄
13. 𝒆(𝒂𝒙+𝒃)
𝒅𝒙 =
𝟏
𝒂
𝒆(𝒂𝒙+𝒃)
+ 𝒄
14. 𝒂 𝑨𝒙+𝑩 𝒅𝒙 =
𝟏
𝑨
𝒂 𝑨𝒙+𝑩
𝐥𝐨𝐠 𝒆 𝒂
+ 𝒄
4. Integration of composite function Part - 2

11. definite integral of 𝒚 = 𝒇(𝒙) between 𝒙 = 𝒂 𝒂𝒏𝒅 𝒙 = 𝒃 is
Given as
Let 𝒇 𝒙 𝒅𝒙 = 𝑭 𝒙 + 𝒄 then
𝒂
𝒃
𝒇(𝒙) 𝒅𝒙 = 𝒗𝒂𝒍𝒖𝒆 𝒐𝒇 𝑭 𝒙 + 𝒄 𝒂𝒕 𝒙 = 𝒃 − 𝒗𝒂𝒍𝒖𝒆 𝒐𝒇 𝑭 𝒙 + 𝒄 𝒂𝒕 𝒙 = 𝒂
= [𝑭 𝒃 + 𝒄]-[𝑭 𝒂 + 𝒄]
= 𝑭 𝒃 + 𝒄 − 𝑭 𝒂 − 𝒄
= 𝑭 𝒃 − 𝑭(𝒂)
= 𝑭(𝒙) 𝒂
𝒃
5. Definition of definite integral

12. 1. 𝒂
𝒃
𝒇(𝒙) 𝒅𝒙= 𝒂
𝒃
𝒇(𝒕) 𝒅𝒕
2. 𝒂
𝒃
𝒇(𝒙) 𝒅𝒙 = – 𝒃
𝒂
𝒇(𝒙) 𝒅𝒙 … [Also, 𝒂
𝒂
𝒇(𝒙) 𝒅𝒙 = 𝟎]
3. 𝒂
𝒃
𝒇(𝒙) 𝒅𝒙 = 𝒂
𝒄
𝒇(𝒙) 𝒅𝒙 + 𝒄
𝒃
𝒇(𝒙) 𝒅𝒙
4. 𝒂
𝒃
𝒇(𝒙) 𝒅𝒙 = 𝒂
𝒃
𝒇(𝒂 + 𝒃 − 𝒙) 𝒅𝒙
5. 𝟎
𝒂
𝒇(𝒙) 𝒅𝒙 = 𝟎
𝒂
𝒇(𝒂 − 𝒙) 𝒅𝒙]
6. 𝟎
𝟐𝒂
𝒇(𝒙) 𝒅𝒙 = 𝟎
𝒂
𝒇(𝒙) 𝒅𝒙 + 𝟎
𝒂
𝒇(𝟐𝒂 − 𝒙) 𝒅𝒙
6. Properties of definite integral with simple problems -1

13. 7. Two parts
𝟎
𝟐𝒂
𝒇(𝒙) 𝒅𝒙 = 𝟐 𝟎
𝒂
𝒇(𝒙) 𝒅𝒙 … if f(2a – x) = f(x).
𝟎
𝟐𝒂
𝒇(𝒙) 𝒅𝒙 = 𝟎 … if f(2a – x) = – f(x)
8. Two parts
−𝒂
𝒂
𝒇(𝒙) 𝒅𝒙 = 𝟐. 𝟎
𝒂
𝒇(𝒙) 𝒅𝒙 … if f(- x) = f(x) or it is an even function
−𝒂
𝒂
𝒇(𝒙) 𝒅𝒙 = 𝟎 … if f(- x) = – f(x) or it is an odd function
6. Properties of definite integral with simple problems -2

14. The area between the graph of y = f(x) and the x-axis is given by the definite
integral below. This formula gives a positive result for a graph above the x-axis, and
a negative result for a graph below the x-axis.
Note: If the graph of y = f(x) is partly above and partly below the x-axis, the
formula given below generates the net area. That is, the area above the axis minus
the area below the axis.
7. Area under the curve

15. Find the area between y = 7 – x2 and the x-axis between the values x = –1 and x = 2.
Example of Area under the curve

16. The area between the two curves or function is defined as the definite
integral of one function (say f(x)) minus the definite integral of other
functions (say g(x)).
Thus, it can be represented as the following:
Area between two curves = 𝒂
𝒃
𝒇 𝒙 − 𝒈(𝒙) 𝒅𝒙
8. Area bounded by two curves

17. Case 1:
Consider two curves y=f(x) and y=g(x), where f(x) ≥ g(x) in [a, b]. In
the given case, the point of intersection of these two curves can be
given as x=a and x=b
Case 2:
Consider another case, when two curves y=f(x) and y=g(x) are given,
such that f(x) ≥ g(x) between x=a and x=c and f(x) ≤ g(x) between x=c
and x=b, as shown in the figure
Total area= 𝑎
𝑐
𝑓 𝑥 − 𝑔(𝑥) 𝑑𝑥 + 𝑐
𝑏
𝑓 𝑥 − 𝑔(𝑥) 𝑑𝑥
How to Find the Area between Two Curves?
A= 𝒂
𝒃
𝒇 𝒙 − 𝒈(𝒙) 𝒅𝒙

18. Find the area of the region bounded by the parabolas y=𝒙 𝟐
and x=𝒚 𝟐
.
= 𝟎
𝟏
𝒇 𝒙 − 𝒈(𝒙) 𝒅𝒙
= 𝟎
𝟏
𝒙 − 𝒙 𝟐
𝒅𝒙
= 𝟎
𝟏
𝒙
𝟏
𝟐 𝒅𝒙 − 𝟎
𝟏
𝒙 𝟐
𝒅𝒙
=
𝒙
𝟏
𝟐
+𝟏
𝟏
𝟐
+𝟏
𝟎
𝟏
−
𝒙 𝟐+𝟏
𝟐+𝟏
𝟎
𝟏
=
𝒙
𝟑
𝟐
𝟑
𝟐
𝟎
𝟏
−
𝒙 𝟑
𝟑
𝟎
𝟏
=
𝟐
𝟑
𝟏
𝟑
𝟐 − 𝟎
𝟑
𝟐 −
𝟏
𝟑
𝟏 𝟑
− 𝟎 𝟑
=
𝟐
𝟑
−
𝟏
𝟑
=
𝟏
𝟑
𝒔𝒒. 𝒖𝒏𝒊𝒕𝒔
Example Area bounded by two curves
Hence, Area
between
two curves
Since required area lies between
(0, 0) and (1, 1) Here a=0 and b=1