Revision on
Integration
1
Important Rules
2
Important Rules
න 𝒌 𝒅𝒙 = 𝒌𝒙 + 𝒄 ; 𝒘𝒉𝒆𝒓𝒆 𝒌 & 𝒄 𝒂𝒓𝒆 𝒄𝒐𝒏𝒔𝒕𝒂𝒏𝒕𝒔
𝒏+𝟏
𝒙
න 𝒙 𝒏 𝒅𝒙 =
+ 𝒄 ; 𝒘𝒉𝒆𝒓𝒆 𝒏
𝒏+𝟏
𝒊𝒔 𝒂 𝒄𝒐𝒏𝒔𝒕𝒂𝒏𝒕 𝒂𝒏𝒅 ≠ −𝟏
න 𝒇′ 𝒙 𝒇(𝒙)
𝒏
𝒅𝒙 =
𝒇(𝒙) 𝒏+𝟏
+ 𝒄 ; 𝒘𝒉𝒆𝒓𝒆 𝒏
𝒏+𝟏
𝒊𝒔 𝒂 𝒄𝒐𝒏𝒔𝒕𝒂𝒏𝒕 𝒂𝒏𝒅 ≠ −𝟏
න 𝒇′ 𝒙 𝒆 𝒇(𝒙) 𝒅𝒙 = 𝒆 𝒇(𝒙) + 𝒄
𝒆𝒈: න
;
𝟏
𝒅𝒙 = 𝒍𝒏 𝒙 + 𝒄
𝒙
න 𝒖 𝒅𝒗 = 𝒖 𝒗 − න 𝒗 𝒅𝒖 ;
;
𝒆𝒈: න 𝒄𝒐𝒔 𝒙 𝒆 𝒔𝒊𝒏 𝒙 𝒅𝒙 = 𝒆 𝒔𝒊𝒏 𝒙
𝒇′ 𝒙
න
𝒅𝒙 = 𝒍𝒏 𝒇(𝒙) + 𝒄
𝒇(𝒙)
+𝒄
𝒆𝒈: න 𝒙 𝒆 𝒙 𝒅𝒙 = 𝒙 𝒆 𝒙 − න 𝒆 𝒙 𝒅𝒙 = 𝒙 𝒆 𝒙 − 𝒆 𝒙 + 𝒄
3
Important Rules
4
Important Rules
5
Examples on Integration
∗ න 𝑇𝑎𝑛 𝑥 𝑑𝑥 =
6
Examples on Integration
∗ න 𝑇𝑎𝑛2 𝑥 𝑑𝑥 =
න( 𝑆𝑒𝑐 2 𝑥 − 1 ) 𝑑𝑥 =
∗ න 𝑥 2 𝑒 𝑥 𝑑𝑥 =
Integration by
Parts
7
Examples on Integration
You may also use DI method in solving similar examples:
∗ න 𝑥 2 𝑒 −𝑥 𝑑𝑥 =
8
Examples on Integration
∗ න 𝑦 𝑠𝑖𝑛 𝑦 𝑑𝑦 =
1
∗ න ln 𝑥 𝑑𝑥 = 𝑙𝑛𝑥 𝑥 − න 𝑥
𝑑𝑥 = 𝑥𝑙𝑛 𝑥 − න 1 𝑑𝑥 =
𝑥
𝑥𝑙𝑛 𝑥 − 𝑥 + 𝑐
9
Examples on Integration
∗ න 𝑒 𝑡 𝑠𝑖𝑛 𝑡 𝑑𝑡 =
Chain Rule
10
Examples on Integration
∗ න𝑥 𝑒
𝑥2
1
1 𝑥2
𝑥2
𝑑𝑥 = න 2 𝑥 𝑒 𝑑𝑥 = 𝑒
+𝑐
2
2
𝑠𝑖𝑛2 𝑥
∗ න sin 𝑥 cos 𝑥 𝑑𝑥 =
+𝑐
2
𝑂𝑅
−𝑐𝑜𝑠 2 𝑥
+𝑐
2
4
−𝑐𝑜𝑠
𝑥
3
∗ න sin 𝑥 𝑐𝑜𝑠 𝑥 𝑑𝑥 =
+𝑐
4
−2
2
sin 𝑥
−𝑐𝑜𝑠
𝑥
1
𝑠𝑒𝑐
𝑥
−3
∗න
𝑑𝑥 = න sin 𝑥 𝑐𝑜𝑠 𝑥 𝑑𝑥 =
+𝑐 =
+𝑐 =
+𝑐
3
2
𝑐𝑜𝑠 𝑥
−2
2𝑐𝑜𝑠 𝑥
2
11
Examples on Integration
sin 𝑥
∗න
𝑑𝑥 =
1 − sin 𝑥
sin 𝑥
=න
𝑑𝑥
1 − sin 𝑥
∗
𝟏 + 𝒔𝒊𝒏 𝒙
sin 𝑥 + sin2 𝑥
= න
𝑑𝑥
𝟏 + 𝒔𝒊𝒏 𝒙
1 − sin2 𝑥
sin 𝑥 + sin2 𝑥
=න
𝑑𝑥 =
cos2 𝑥
+
sin2 𝑥
න
𝑑𝑥
cos2 𝑥
+
න tan2 𝑥 𝑑𝑥
= − න − sin 𝑥 (cos 𝑥 )−2 𝑑𝑥
+
න(sec 2 𝑥 − 1 ) 𝑑𝑥
(cos 𝑥 )−1
=−
−1
+
tan 𝑥 − 𝑥 + 𝑐
=
sin 𝑥
න
𝑑𝑥
cos2 𝑥
න sin 𝑥 (cos 𝑥 )−2 𝑑𝑥
= (cos 𝑥 )−1 +
tan 𝑥 − 𝑥 + 𝑐
12
Examples on Integration
1
∗න
𝑑𝑡 =
−𝑡
1+ 𝑒
13
Partial Fractions and Long Division
Partial fractions:
We use partial fractions when the highest power of a variable in the
numerator is less than the highest power of which in the denominator .
Long division:
We use long division when the highest power of a variable in the
numerator is greater than or equal to the highest power of which in the
denominator.
14
Examples on Partial Fraction Method
4
∗න
𝑑𝑥 =
𝑥 + 2 (𝑥 + 1)
Case # 1
15
Examples on Partial Fraction Method
4
∗න
𝑑𝑥
𝑥 + 2 (𝑥 + 1)
16
Examples on Partial Fraction Method
1
∗න
𝑑𝑥 =
2
𝑥 + 2 (𝑥 − 3)
Case # 2
1/25
-1/25
1/5
17
Examples on Partial Fraction Method
Case # 2
1
A
∗න
𝑑𝑥 ≡ න
2
𝑥 + 2 (𝑥 − 3)
𝑥+2
1/25
≡ න
𝑥+2
1
1
≡
න
25
𝑥+2
1
≡
ln 𝑥 + 2
25
≡ 𝐥𝐧
𝑑𝑥
B
+න
𝑥−3
𝑑𝑥
C
+න
𝑑𝑥
2
(𝑥 − 3)
𝑑𝑥
−1/25
+න
𝑑𝑥
𝑥−3
1/5
+න
𝑑𝑥
2
(𝑥 − 3)
1
1
1
𝑑𝑥 − න
𝑑𝑥 + න(𝑥 − 3)−2 𝑑𝑥
25
𝑥−3
5
1
1
−
ln 𝑥 − 3
− 𝑥 − 3 −1 + 𝑐
25
5
𝒙+𝟐
𝒙−𝟑
𝟏
𝟐𝟓
𝟏
𝟓
−
+𝑪
𝒙−𝟑
18
Examples on Partial Fraction Method
1
∗න
𝑑𝑥 =
2
𝑥 + 1 (𝑥 +2𝑥 + 2)
Case # 3
1
A
B𝑥+C
≡
+
𝑥 + 1 (𝑥 2 +2𝑥 + 2)
𝑥+1
(𝑥 2 +2𝑥 + 2)
By multiplying both sides by (x+1)(x2 + 2x + 2) ; therefore
1 ≡ A (x2 + 2x + 2) + (Bx + C) (x+1)
1 ≡ A (x2 + 2x + 2) + (Bx2 + Bx + Cx + C)
By equating coefficients both sides:
coeff. of x2
→
0 ≡ A+ B
coeff. of x
→
0 ≡ 2A + B + C
coeff. of x0 (constant) →
1 ≡ 2A + C
By solving the 3 equations :
A = 1 ; B = -1 ; C = -1
19
Examples on Partial Fraction Method
Case # 3
1
A
B𝑥+C
∗න
𝑑𝑥 = න
𝑑𝑥 + න 2
𝑑𝑥
𝑥 + 1 (𝑥 2 +2𝑥 + 2)
𝑥+1
(𝑥 +2𝑥 + 2)
1
−𝑥 − 1
=න
𝑑𝑥 + න 2
𝑑𝑥
𝑥+1
(𝑥 +2𝑥 + 2)
1
𝑥+1
=න
𝑑𝑥 − න 2
𝑑𝑥
𝑥+1
(𝑥 +2𝑥 + 2)
1
1
2(𝑥 + 1)
=න
𝑑𝑥 − න 2
𝑑𝑥
𝑥+1
2 (𝑥 +2𝑥 + 2)
1
= ln 𝑥 + 1 − ln 𝑥 2 + 2𝑥 + 2 + 𝑐
2
= ln
(𝑥 + 1)
𝑥 2 + 2𝑥 + 2
1
2
+𝑐
20
Examples on Long Division Method
(𝑥 3 + 2)
∗න 2
𝑑𝑥 =
(𝑥 +1)
21
Examples on Long Division Method
(𝑥 3 + 2)
∗න 2
𝑑𝑥 =
(𝑥 +1)
න𝑥
+
−𝑥 3 + 2
𝑥 2 +1
𝑑𝑥
22
23