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Integration by Parts
Motivation: Via Fundamental Theorem of Calculus, the relationship between derivative function and integral is established. Thus, we can collect some properties from derivative functions and apply them while working on the integral. The derivative of the product of two functions is shown as below
(uv) = v + u
Correspondingly, we will have the following for integral:
= + + C1 , C1 is a constant. uv = + + C
It has been explained that the notation, , originally was just represented the derivative function with respect to the variable x; but later on, we figure out that it is the quotient of the small variations on the variable u and x. From that , we have
With the result, we will have simpler representation :
uv = + + C
In general, we can use
= vu - + C
This technique is known as “Integration by Parts”. And we state the whole theorem below.
Theorem ( Integration by Parts ) Let u and v be differentiable functions of x on the interval [a,b] . Then
= - or = - where and are derivative function of u and v respectively.
Example: for x > 0 Sol: =
So, = x ln(x) - + C = x ln(x) - + C = x ln(x) - + C = x ln(x) – x + C
Example: Sol: = sin(x) = (-cos(x))x3 - = -cos(x)x3 +
And we can use integration by parts on again. Integration by parts will be performed many times for this problem. Our final goal is to have those sin(x) or cos(x) term in the integral without being multiplied by the polynomial term. The polynomial term can be eliminated via the differentiation process in “integration by parts”.
We can summarize all the work with the following diagram:
Differentiation Integration x3 sin(x) 3x2 -cos(x)
6x -sin(x) 6 cos(x) 0 sin(x)
The list above is just differentiating x3 all the way down and integrating sin(x) all the way down. With the list, we can write down the result very quickly:
= x3(-cos(x)) – 3x2(-sin(x)) + 6xcos(x) - = x3(-cos(x)) – 3x2(-sin(x)) + 6xcos(x) – 6sin(x) + C |
