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Various Forms and Results of Integrals

Various Forms and Results of Integrals

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Integration by Parts


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

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

Integration by Parts

plain and simple summary

nub

plain and simple summary

nub

dummy

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

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simple steps to build the foundation

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


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Starting on learning "". In this page, integration by parts is explained with examples.

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
  • 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
  • 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` 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
  • 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 `

comprehensive information for quick review

Jogger

comprehensive information for quick review

Jogger

dummy

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`



           

practice questions to master the knowledge

Exercise

practice questions to master the knowledge

Exercise

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

  • `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 `

Integrate `int e^x cos x dx`

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

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

Integrate `int x sin x dx`

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

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

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Integration of functions that are product of sub functions is hard to fund result for. For example, f of x = x squared, and g of x = sine x, can be individually integrated. But integrating product of f of x and g of x, = , x squared, multiplied, sine x, is to be worked out. Knowing that integration is anti-derivative, let us examine how product of sub-functions is differentiated. ;; u v whole prime, =, v, u prime, +, u, v prime. If one of the functions is in x power n form, then repeated differentiation will result in a constant. Knowing this, the equation above is modified to. u v prime = u v whole prime, minus, v u prime. Integrating this equation gives. integral f of x, g of x dx, =, f of x. integral g of x d x, minus, integral, integral g of x d x, d by d x of f of x, dx. ;; The advantage is, the function f of x is differentiated in the second term. If f of x = x, then the second term is simplified to a standard result.
integral f of x, g of x dx, =, f of x. integral g of x d x, minus, integral, integral g of x d x, d by d x of f of x, dx. this is called integration by parts. Note that the integrand is split into two parts.
Integration for x squared sine x dx is attempted. How to proceed further in integration.
integration;parts;can;used;second;term
The integration by parts can be used on the second term too
no;further;steps;possible
no further steps are possible to integrate
The answer is "The integration by parts can be used on the second term too". The rest of integration is given.
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 power x, then as the sequential integration or differentiation does not change e power x. This leads to an interesting form of integration. Integral e power x sine x dx is attempted. The result has the integral we started off with. How to resolve this.
handle;this;equation;and;rearrange
handle this as an equation and rearrange such that the term is in one side of the equation
no;further;steps;possible
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".
Integral e power x sine x dx is given. What is the result.
equals;power;cos;sine;over
I equals e power x cos x + e power x sine x, over 2 + c
canceled;both;sides
I is canceled on both the sides
The answer is "I equals e power x cos x + e power x sine x, over 2 + c"
Another interesting observation in integration by parts: If one of the functions is e power x, then as the sequential integration or differentiation does not change e power x. This leads to an interesting form of integration when the other function is in the form h of x + h prime of x. ;; Integration of that is attempted. Is it possible to simplify this further.
yes;s;cancel;second;term
yes, cancel the second term and third term
no;further;steps;possible
no further steps are possible to integrate
The answer is "yes, cancel the second term and third term"
Integrating integral e power x, multiplied, natural log x + 1 by x, dx is given.
Simplified nub of Integration by parts: Integrand is taken as product of two functions to integrate.
Reference jogger of Integration by Parts: integral f of x, g of x dx, = f of x, integral g of x minus integral, integral g of x dx, d by d x f of x dx. ;; e power x sine x, or, e power 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 power x in Integration by Parts: integral e power x, multiplied, h of x + h prime of x, dx, = , e power x, h of x, + c
integrate the given function
1
2
The answer is "e power x arc tan x + c"
integrate the given function
1
2
The answer is "e power x sine x + cos x, by 2 + c"
integrate the given function
1
2
The answer is "minus x cos x + sine x + c"

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