Table of Contents
Integration by Parts Formula
Integration is one of calculus part. Integration is not only consisting of general formula, but also integration by substitution and integration by parts.
Different problem needs different way to solve it. Integration by parts was discovered by mathematicians, Brook Taylor, in 1715.
Integration by parts is used to solve integration problems that can not be solved by general integration and integration by substitution formula.
In other words, integration by parts’ level is higher than general integration. Integration by parts can be used to solve integration problem that two function are multiplied each other. The formula is:
ʃ u dv = u.v – ʃ v du
Where:
- u: function u(x)
- v: function v(x)
- dv: derivative of v(x)
- du: derivative of u(x)
note: how to determine which one is u and v?
choose the simple function as u, and more complicated as v. in another words, if a function can be derivate in the first, second, and so on until it will get 0, it is u.
On the contrary, if a function can not be derivate until it gets 0, it must be integrated, and it is v.
The formula of integration by parts is not too long and easy to remember, but sometimes it needs more time and effort to determine the solution of integration by parts’ problem.
Integration by part has another way to solve the problem. It simpler than before. It still uses derivative concept and integration concept. Making table will make it easy.
| Derivative of u(x) | Integration of dv |
| u(x) | dv |
| u’(x) | Integration of dv |
| u’’(x) | … |
| …. | …. |
After making the table, the next step is multiplication between first bar of derivative (u(x) value) with first integration of dv.
It has positive sign (+). And next, multiply second bar of derivative of u(x) (first derivative of u(x)) with second integration of dv.
It has negative sign (-). It is continued until derivative’s part can not be derivate to the next bar. It will clearer if you look and learn the example part.
Integration by Parts Examples
- Solve the problem
ʃ 2x.ex = . . . .
Let
u = 2x → du = u’=2
dv = ex → v = ex
Then
ʃ 2x.ex = 2x.ex – ʃ ex.2dx
= 2x.ex – 2 ʃ exdx
= 2xex – 2ex + c
= 2xex (x-1) + c
Another way to solve the problem
| 2x | ex |
| 2 | ex |
| 0 | ex |
Next step is multiplication that is like the rules
2x.ex – 2(ex) = 2.ex(x-1)
2. Solve the problem
ʃ (3t + 5).cos (t/4) = . . . .
Let
u = 3t+5 → u’ = 3
dv = cos (t/4) → v = 4sin(t/4)
Then
ʃ (3t + 5).cos (t/4) = (3t + 5).4.sin (t/4) – ʃ 4. sin (t/4). 3dt
= 4(3t + 5).sin (t/4) – 12ʃ sin (t/4) dt
= 4(3t + 5).sin (t/4) – 46 cos (t/4) + c
Using another way
| 3t+5 | Cos(t/4) |
| 3 | 4.Sin(t/4) |
| 0 | 16.cos(t/4) |
Be careful about the sign and rule. It will become:
(3t+5).4.sin(t/4) – 3.16.cos(t/4) = 4.(3t+5).sin(t/4) – 48. Cos(t/4)