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Thought-Process to Discover Knowledge

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Books and other education websites provide "matter-of-fact" knowledge. Instead, nubtrek provides a thought-process to discover knowledge.

In each of the topic, the outline of the thought-process, for that topic, is provided for learners and educators.

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In this page, integration by parts is explained with examples.



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Integration of functions that are product of sub-functions is hard to find result for. For example, `f(x)=x^2` and `g(x)=sinx` can be individually integrated. But integrating `f(x)g(x) = x^2 sinx` is to be worked out.

Knowing that integration is anti-derivative, let us examine how product of sub-functions is differentiated.
`(uv)′ = vu′+uv′`

If one of the functions is in `x^n` form, then repeated differentiation will result in a constant. Knowing this, the equation above is modified to
`u v′ = (uv)′ - vu′ `

integrating this
`int u v′ dx = int (uv)′ dx - int v u′ dx `

`int u v′ dx = uv - int v u′ dx `

substituting `u= f(x)` and `v′ = g(x)` and so `v= int g(x) dx`

`int f(x) g(x) dx ` `= f(x) intg(x)dx ``- int [int g(x) dx] [d/(dx)f(x)] dx`

The advantage is, the function `f(x)` is differentiated in the second term. If `f(x)=x` then the second term is simplified to a standard result.

`int f(x) g(x) dx ` `= f(x) intg(x)dx ``- int [int g(x) dx] [d/(dx)f(x)] dx`

This is called integration by parts. Note that the integrand is split into two parts.

`int x^2 sin x dx`

using integration by parts
`=-x^2cos x - int (-cos x) (2x)dx+c`

`=-x^2cos x + int 2xcos x dx+c`

How to proceed further in integration?

  • The integration by parts can be used on the second term too
  • The integration by parts can be used on the second term too
  • no further steps are possible to integrate

The answer is "The integration by parts can be used on the second term too"

`=-x^2cos x + int 2xcos x dx+c`

using integration by parts on the second term
`=-x^2cos x + 2x sin x - int sinx xx2 dx+ c `

`= -x^2cos x + 2x sin x + 2cos x+ c`

Interesting observation in the integration by parts is that one of the two sub-functions is differentiated sequentially. The other is integrated repeatedly. If one of the functions is `e^x`, then as the sequential integration or differentiation does not change `e^x`. This leads to an interesting form of integration.

`int e^x sin x dx `

`= e^x(-cos x)``+ int e^x cos x dx + c`

`= -e^xcos x + e^x sinx ``- int e^x sin x dx + c`

The result has the integral we started off with. How to resolve this?

  • handle this as an equation and rearrange such that the term is in one side of the equation
  • handle this as an equation and rearrange such that the term is in one side of the equation
  • no further steps are possible to integrate

The answer is "handle this as an equation and rearrange such that the term is in one side of the equation".

`I = int e^x sin x dx `

`= e^x(-cos x)``+ int e^x cos x dx + c`

`= -e^xcos x + e^x sinx ``- int e^x sin x dx + c`

The above is written as `I = -e^xcos x + e^x sinx - I + c`

what is the result `I`?

  • `I=(-e^xcos x + e^x sinx)//2 + c`
  • `I=(-e^xcos x + e^x sinx)//2 + c`
  • `I` is canceled on both the sides

The answer is "`I=(-e^xcos x + e^x sinx)//2 + c`".

Another interesting observation in integration by parts: If one of the functions is `e^x`, then as the sequential integration or differentiation does not change `e^x`. This leads to an interesting form of integration when the other function is in the form. `h(x)+h′(x)`.

`int e^x (h(x)+h′(x)) dx `

`= int e^x h(x) dx ``+ int e^x h′(x) dx `

applying integration by parts in the first integral

`= e^x h(x) ``- int e^x h′(x) dx + int e^x 1/x dx +c`

Is it possible to simplify this further?

  • yes, cancel the second term and third term
  • yes, cancel the second term and third term
  • no further steps are possible to integrate

The answer is "yes, cancel the second term and third term". The integration result is `e^xh(x) + c`.

Integrate `int e^x (ln x + 1/x) dx`.

`int e^x (ln x + 1/x) dx `

`= int e^x ln x dx + int e^x 1/x dx ` `= e^x ln x ``- int e^x 1/x dx + int e^x 1/x dx +c` `= e^x ln x +c `

Integration by Parts: Integrand is taken as product of two functions to integrate.

Integration by Parts: `int f(x) g(x) dx ` `= f(x) int g(x)dx - int [int g(x) dx] [d/(dx)f(x)] dx`

`e^x sin x` or `e^x cos x` in Integration by Parts: After two iterations of integration by parts, the integral is solved by rearranging the terms of the equation.

`e^x` in Integration by Parts: `int e^x (h(x)+h′(x)) dx ``=e^x h(x) + c`

Solved Exercise Problem:

Integrate `int e^x arctan x + e^x/(1+x^2) dx`

  • `e^x arctan x + c`
  • `e^x arctan x + c`
  • `e^x/(1+x^2) + c`

The answer is "`e^x arctan x + c`". The integral is in the form `int e^x (h(x)+h′(x)) dx `

Solved Exercise Problem:

Integrate `int e^x cos x dx`

  • `e^x (sinx -cos x)//2 + c`
  • `e^x (sin x+cos x)//2 + c`
  • `e^x (sin x+cos x)//2 + c`

The answer is "`e^x (sin x+cos x)//2 + c`".

Solved Exercise Problem:

Integrate `int x sin x dx`

  • `-x cos x + sinx +c`
  • `-x cos x + sinx +c`
  • `x cos x + sin x + c`

The answer is "`-x cos x + sinx +c`".

                            
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