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Integration by Parts

Integration by Parts is a special method of integration that is often useful when two functions are multiplied together, but is also helpful in other ways.

You will see plenty of examples soon, but first let us see the rule:
Integration by parts method, Business Mathematics & Statistics | Business Mathematics and Statistics - B Com

  • u is the function u(x)
  • v is the function v(x)

As a diagram
Integration by parts method, Business Mathematics & Statistics | Business Mathematics and Statistics - B Com
 Let's get straight into an example, and talk about it after:

Example: What is Integration by parts method, Business Mathematics & Statistics | Business Mathematics and Statistics - B Com ?

OK, we have x multiplied by cos(x), so integration by parts is a good choice.

First choose which functions for u and v:

  • u = x
  • v = cos(x)

So now it is in the format Integration by parts method, Business Mathematics & Statistics | Business Mathematics and Statistics - B Comwe can proceed:

Differentiate u: u' = x' = 1

Integrate vIntegration by parts method, Business Mathematics & Statistics | Business Mathematics and Statistics - B Com   (see Integration Rules)

Now we can put it togethe

Integration by parts method, Business Mathematics & Statistics | Business Mathematics and Statistics - B Com

 

Simplify and solve:

x sin(x) − ∫sin(x) dx

x sin(x) + cos(x) + C

So we followed these steps:

  • Choose u and v
  • Differentiate u: u'
  • Integrate v:Integration by parts method, Business Mathematics & Statistics | Business Mathematics and Statistics - B Com
  • Put u, u' and ∫v dx into:u Integration by parts method, Business Mathematics & Statistics | Business Mathematics and Statistics - B Com
  • Simplify and solve

In English, to help you remember, Integration by parts method, Business Mathematics & Statistics | Business Mathematics and Statistics - B Combecomes:

(u integral v) minus integral of (derivative u, integral v)

Let's try some more examples:

Example: What is  Integration by parts method, Business Mathematics & Statistics | Business Mathematics and Statistics - B Com

First choose u and v:

  • u = ln(x)
  • v = 1/x2

Differentiate u: ln(x)' = 1/x

Integrate v: Integration by parts method, Business Mathematics & Statistics | Business Mathematics and Statistics - B Com  (by the power rule)

Now put it together:

Integration by parts method, Business Mathematics & Statistics | Business Mathematics and Statistics - B Com

Simplify:

− ln(x)/x − Integration by parts method, Business Mathematics & Statistics | Business Mathematics and Statistics - B Com

− (ln(x) + 1)/x + C

 Example: What is Integration by parts method, Business Mathematics & Statistics | Business Mathematics and Statistics - B Com

But there is only one function! How do we choose u and v ?

Hey! We can just choose v as being "1":

  • u = ln(x)
  • v = 1

Differentiate u: ln(x)' = 1/x

Integrate v:Integration by parts method, Business Mathematics & Statistics | Business Mathematics and Statistics - B Com

Now put it together:

Integration by parts method, Business Mathematics & Statistics | Business Mathematics and Statistics - B Com

 

Integration by parts method, Business Mathematics & Statistics | Business Mathematics and Statistics - B Com

 Example: What is   Integration by parts method, Business Mathematics & Statistics | Business Mathematics and Statistics - B Com

Choose u and v:

  • u = ex
  • v = x

Differentiate u: (ex)' = ex

Integrate Integration by parts method, Business Mathematics & Statistics | Business Mathematics and Statistics - B Com

Now put it together: 

Integration by parts method, Business Mathematics & Statistics | Business Mathematics and Statistics - B Com

Well, that was a spectacular disaster! It just got more complicated.

Maybe we could choose a different u and v?
 Example:   Integration by parts method, Business Mathematics & Statistics | Business Mathematics and Statistics - B Com

Choose u and v differently:

  • u = x
  • v = ex

Differentiate u: (x)' = 1
 Integrate   Integration by parts method, Business Mathematics & Statistics | Business Mathematics and Statistics - B Com

Now put it together:

Integration by parts method, Business Mathematics & Statistics | Business Mathematics and Statistics - B Com

Simplify:

x ex − ex + C

ex(x−1) + C

The moral of the story: Choose u and v carefully!

Choose a u that gets simpler when you differentiate it and a v that doesn't get any more complicated when you integrate it.

A helpful rule of thumb is I LATE. Choose u based on which of these comes first:

  • I: Inverse trigonometric functions such as sin-1(x), cos-1(x), tan-1(x)
  • L: Logarithmic functions such as ln(x), log(x)
  • A: Algebraic functions such as x2, x3
  • T: Trigonometric functions such as sin(x), cos(x), tan (x)
  • E: Exponential functions such as ex, 3x

And here is one last (and tricky) example:
Example: Integration by parts method, Business Mathematics & Statistics | Business Mathematics and Statistics - B Com

  • u = sin(x)
  • v = ex

Differentiate u: sin(x)' = cos(x)
 Integrate   Integration by parts method, Business Mathematics & Statistics | Business Mathematics and Statistics - B Com

Now put it together:
Integration by parts method, Business Mathematics & Statistics | Business Mathematics and Statistics - B Com    

Simplify:

Integration by parts method, Business Mathematics & Statistics | Business Mathematics and Statistics - B Com

Now we have the same integral on both sides (except one is subtracted) ...

... so bring the right hand one over to the left and we get:
Integration by parts method, Business Mathematics & Statistics | Business Mathematics and Statistics - B Com
 Simplify:
Integration by parts method, Business Mathematics & Statistics | Business Mathematics and Statistics - B Com

Footnote: Where Did "Integration by Parts" Come From?

It is based on the Product Rule for Derivatives:

(uv)' = uv' + u'v

Integrate both sides and rearrange:

Integration by parts method, Business Mathematics & Statistics | Business Mathematics and Statistics - B Com 

Some people prefer that last form, but I like to integrate v' so the left side is simple:

Integration by parts method, Business Mathematics & Statistics | Business Mathematics and Statistics - B Com

The document Integration by parts method, Business Mathematics & Statistics | Business Mathematics and Statistics - B Com is a part of the B Com Course Business Mathematics and Statistics.
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FAQs on Integration by parts method, Business Mathematics & Statistics - Business Mathematics and Statistics - B Com

1. What is the integration by parts method?
Ans. The integration by parts method is a technique used in calculus to evaluate integrals that are in the form of the product of two functions. It is based on the product rule for differentiation and involves choosing one function to differentiate and another function to integrate.
2. How does the integration by parts method work?
Ans. The integration by parts method works by applying the formula: ∫u dv = uv - ∫v du, where u and v are functions of x. The goal is to choose u and dv in such a way that the integral on the right side becomes easier to solve than the original integral.
3. When should I use the integration by parts method?
Ans. The integration by parts method is most useful when you have an integral that is the product of two functions and you are unable to evaluate it using other integration techniques such as substitution or trigonometric identities. It is particularly helpful when one of the functions becomes simpler when differentiated or integrated.
4. Can you provide an example of using the integration by parts method?
Ans. Sure! Let's say we want to evaluate the integral of x * e^x dx. We can choose u = x and dv = e^x dx. Then, du = dx and v = ∫e^x dx = e^x. Applying the integration by parts formula, we have: ∫x * e^x dx = x * e^x - ∫e^x dx Simplifying further, we get: ∫x * e^x dx = x * e^x - e^x + C where C is the constant of integration.
5. Are there any limitations or conditions for using the integration by parts method?
Ans. Yes, there are certain conditions for using the integration by parts method. One important condition is that the integral should be able to be split into a product of two functions. Additionally, the functions involved should be differentiable and integrable, meaning they should have well-defined derivatives and antiderivatives. If these conditions are not met, the integration by parts method may not be applicable.
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