Integration by
Parts
Example:
� 𝑥𝑥 𝑒𝑒 𝑥𝑥 𝑑𝑑𝑑𝑑
𝒖𝒖 = 𝒙𝒙 𝒅𝒅𝒅𝒅 = 𝒆𝒆𝒙𝒙 𝒅𝒅𝒅𝒅
𝑑𝑑𝑑𝑑 = 𝑑𝑑𝑑𝑑 𝑣𝑣 = 𝑒𝑒 𝑥𝑥
𝑥𝑥 𝑥𝑥 𝑥𝑥
∫ 𝑥𝑥 𝑒𝑒 𝑑𝑑𝑑𝑑 = 𝑥𝑥𝑒𝑒 − ∫ 𝑑𝑑𝑑𝑑
𝑒𝑒
= 𝑥𝑥𝑒𝑒 𝑥𝑥 − 𝑒𝑒 𝑥𝑥 + 𝐶𝐶
Steps:
1- Determine the 𝑢𝑢 value by using LIPET rule
• L = Logarithmic function
• I = Inverse trigonometric function
• P =Polynomial function
• E = Exponential function
• T = Trigonometric function
2- Let 𝑑𝑑𝑑𝑑 be other function
3- Find 𝑑𝑑𝑑𝑑 and 𝑣𝑣
4- Plug into formula and integrate
� sin−1 𝑥𝑥 𝑑𝑑𝑑𝑑
𝑢𝑢 = sin−1 𝑥𝑥 𝑑𝑑𝑑𝑑 = 𝑑𝑑𝑑𝑑
1 𝑣𝑣 = 𝑥𝑥
𝑑𝑑𝑑𝑑 = 𝑑𝑑𝑑𝑑
1 − 𝑥𝑥 2
𝑥𝑥
∫ sin−1 𝑥𝑥 𝑑𝑑𝑑𝑑 = 𝑥𝑥 sin−1 𝑥𝑥 − ∫ 1−𝑥𝑥 2
𝑑𝑑𝑑𝑑
−1
1
= 𝑥𝑥 sin−1 𝑥𝑥 − ∫ −2𝑥𝑥 1 − 𝑥𝑥 2 2 𝑑𝑑𝑑𝑑
−2
1
1 2
1−𝑥𝑥 2
= 𝑥𝑥 sin−1 𝑥𝑥 + 1 + 𝐶𝐶
2
2
1
= 𝑥𝑥 sin−1 𝑥𝑥 + 1 − 𝑥𝑥 2 2 + 𝐶𝐶
Tabular Method:
Example: + 𝑥𝑥 3 𝑒𝑒 2𝑥𝑥
− 3𝑥𝑥 2 1 2𝑥𝑥
∫ 𝑥𝑥 3 𝑒𝑒 2𝑥𝑥 𝑑𝑑𝑑𝑑 2
𝑒𝑒
1 3 2𝑥𝑥 3 2 2𝑥𝑥 + 6𝑥𝑥 1 2𝑥𝑥
= 𝑥𝑥 𝑒𝑒 − 𝑥𝑥 𝑒𝑒 𝑒𝑒
2 4 4
− 6 1 2𝑥𝑥
6 6 𝑒𝑒
+ 𝑥𝑥 𝑒𝑒 − 𝑒𝑒 2𝑥𝑥 +
2𝑥𝑥
𝐶𝐶 8
8 16
0 1 2𝑥𝑥
𝑒𝑒
16
Example: + 𝑥𝑥 2 cos 3𝑥𝑥
− 2𝑥𝑥 1
∫ 𝑥𝑥 2 cos 3𝑥𝑥 𝑑𝑑𝑑𝑑 3
sin 3𝑥𝑥
1 2 2 + 2 −1
= 𝑥𝑥 sin 3𝑥𝑥 + 𝑥𝑥 cos 3𝑥𝑥 9
cos 3𝑥𝑥
3 9
2 − 0 −1
− sin 3𝑥𝑥 + 𝐶𝐶 27
sin 3𝑥𝑥
27
Example: + 𝑥𝑥 3 + 2𝑥𝑥 sin 2𝑥𝑥
− 3𝑥𝑥 2 +2 −1
∫(𝑥𝑥 3 + 2𝑥𝑥) sin 2𝑥𝑥 𝑑𝑑𝑑𝑑 cos 2𝑥𝑥
2
−1 + 6𝑥𝑥 −1
= (𝑥𝑥 3 + 2𝑥𝑥) cos 2𝑥𝑥 sin 2𝑥𝑥
2 4
1 − 6 1
+ (3𝑥𝑥 2 + 2) sin 2𝑥𝑥 cos 2𝑥𝑥
4 8
6 0 1
+ 𝑥𝑥 cos 2𝑥𝑥 16
sin 2𝑥𝑥
8
6
− sin 2𝑥𝑥 + 𝐶𝐶
16
Repeated Integrals:
Example:
Solution:
Another solution: + 𝑒𝑒 𝑥𝑥 cos 𝑥𝑥
∫ 𝑒𝑒 𝑥𝑥 cos 𝑥𝑥 𝑑𝑑𝑑𝑑 − 𝑒𝑒 𝑥𝑥 sin 𝑥𝑥
𝑥𝑥 𝑥𝑥 +� 𝑒𝑒 𝑥𝑥 − cos 𝑥𝑥
= 𝑒𝑒 sin 𝑥𝑥 + 𝑒𝑒 cos 𝑥𝑥
− ∫ 𝑒𝑒 𝑥𝑥 cos 𝑥𝑥 𝑑𝑑𝑑𝑑
2∫ 𝑒𝑒 𝑥𝑥 cos 𝑥𝑥 𝑑𝑑𝑑𝑑 = 𝑒𝑒 𝑥𝑥 sin 𝑥𝑥 + 𝑒𝑒 𝑥𝑥 cos 𝑥𝑥
𝑥𝑥 1 𝑥𝑥
∫ 𝑒𝑒 cos 𝑥𝑥 𝑑𝑑𝑑𝑑 = [𝑒𝑒 sin 𝑥𝑥 + 𝑒𝑒 𝑥𝑥 cos 𝑥𝑥] + 𝐶𝐶
2
3𝑥𝑥 + 𝑒𝑒 3𝑥𝑥 cos 2𝑥𝑥
Example:∫ 𝑒𝑒 cos 2𝑥𝑥 𝑑𝑑𝑑𝑑 − 3𝑒𝑒 3𝑥𝑥 1
sin 2𝑥𝑥
Solution: 9𝑒𝑒 3𝑥𝑥
2
−1
+� cos 2𝑥𝑥
4
∫ 𝑒𝑒 3𝑥𝑥 cos 2𝑥𝑥 𝑑𝑑𝑑𝑑
1 3𝑥𝑥 3 3𝑥𝑥 9
= 𝑒𝑒 sin 2𝑥𝑥 + 𝑒𝑒 cos 2𝑥𝑥 − ∫ 𝑒𝑒 3𝑥𝑥 cos 2𝑥𝑥 𝑑𝑑𝑑𝑑
2 4 4
13 3𝑥𝑥 1 3𝑥𝑥 3 3𝑥𝑥
4
∫ 𝑒𝑒 cos 2𝑥𝑥 𝑑𝑑𝑑𝑑 = 2 𝑒𝑒 sin 2𝑥𝑥 + 4 𝑒𝑒 cos 2𝑥𝑥
3𝑥𝑥 4 1 3𝑥𝑥 3 3𝑥𝑥
∫ 𝑒𝑒 cos 2𝑥𝑥 𝑑𝑑𝑑𝑑 = 13 [2 𝑒𝑒 sin 2𝑥𝑥 + 4 𝑒𝑒 cos 2𝑥𝑥] + 𝐶𝐶
Example: 𝒖𝒖 = cos ln 𝑥𝑥 𝒅𝒅𝒅𝒅 = 𝒅𝒅𝒅𝒅
Evaluate: ∫ cos ln 𝑥𝑥 𝑑𝑑𝑑𝑑 𝑑𝑑𝑑𝑑 = −sin ln 𝑥𝑥
1
𝑥𝑥
𝑣𝑣 = 𝑥𝑥
Solution:
∫ cos ln 𝑥𝑥 𝑑𝑑𝑑𝑑 = 𝑥𝑥 cos ln 𝑥𝑥 + ∫ sin ln 𝑥𝑥 𝑑𝑑𝑑𝑑
