nubtrek

Server Error

Server Not Reachable.

This may be due to your internet connection or the nubtrek server is offline.

Thought-Process to Discover Knowledge

Welcome to nubtrek.

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.

Read in the blogs more about the unique learning experience at nubtrek.continue

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



click on the content to continue..

Consider the integration `int 2x sin x^2 dx`
We notice that `d/(dx) x^2 = 2x` is part of the integration.

Which of the following can be used to work out the integration?

  • take `t=x^2` and substitute in the integrand
  • take `t=x^2` and substitute in the integrand
  • noticing `d/(dx) x^2 = 2x` does not help

The answer is "take `t=x^2` and substitute in the integrand".

integration `int 2x sin x^2 dx`

Substitute `t=x^2` and `dt = 2xdx`.

`int 2x sin x^2 dx`

`= int sin t dt`

`= -cos t +c`

substitute `t=x^2`
`= -cos x^2 + c`

Considering the integration `int f(x)dx` :
If it is observed that `f(x)` can be rewritten as
`f(x) = h[g(x)] g′(x)`

then,
`int f(x)dx = int h(y)dy`
where `y=g(x)`
and `dy = g′(x)dx`

This method is "integration by substitution".

Integration by substitution : `int h(g(x)) g′(x) dx = [int h(y)dy]|_(y=g(x))`

Integration by Substitution: When `f(x) = h(g(x)) g′(x)`

then,
`int f(x)dx = int h(y)dy`
where `y=g(x)`
and `dy = g′(x)dx` and so `int f(x)dx ``= [int h(y)dy]|_(y=g(x)) + c`

Given a function in parametric form `y=rcos t` and `x=rsin t`, How to proceed with the integration `int ydx`?

  • Substitute `y=r cos t` and `dx = d (r cos t)`. Then, proceed with variable of integration as `t`
  • Substitute `y=r cos t` and `dx = d (r cos t)`. Then, proceed with variable of integration as `t`
  • the function in parametric form cannot be integrated

The answer is 'Substitute `y=r cos t` and `dx = d (r cos t)`'

Given a function in parametric form `y=rcos t` and `x=rsin t`.
`int ydx`

substituting `y=rcos t` and `dx = d(r sin t)`. the `d()` means derivative of
`= int r cost d(r sin t)`

`= int r cos t r cos t dt`

`= int r^2 cos^2 t dt`

This can be integrated.

Solved Exercise Problem:

Integrate `int tan x dx`

  • can be integrated with `tan x = - (cos x)′//cos x`
  • can be integrated with `tan x = - (cos x)′//cos x`
  • no known method to integrate

The answer is "can be integrated"

`int tanx dx`

`=int (- (cos x)′)/(cos x)`

`=-log|cos x| + c`

Solved Exercise Problem:

Integrate `int cot x dx`

  • can be integrated with `cot x = (sin x)′//sin x`
  • can be integrated with `cot x = (sin x)′//sin x`
  • no known method to integrate

The answer is "can be integrated"

`int tanx dx`

`=((sin x)′)/(sin x)`

`=log|sin x| + c`

Solved Exercise Problem:

Integrate `int sec x dx`

  • can be integrated with `sec x ``= (sec x + tan x)′``//(sec x + tan x)`
  • can be integrated with `sec x ``= (sec x + tan x)′``//(sec x + tan x)`
  • no known method to integrate

The answer is "can be integrated"

`int sec x dx`

`=int ((sec x + tan x)′)/(sec x + tan x) dx`

`=log|sec x + tan x|+c `

Solved Exercise Problem:

Integrate `int csc x dx`

  • can be integrated with `csc x ``= -(csc x + cot x)′``//(csc x + cot x)`
  • can be integrated with `csc x ``= -(csc x + cot x)′``//(csc x + cot x)`
  • no known method to integrate

The answer is "can be integrated"

`int csc x dx`

`=int (-(csc x + cot x)′)/(csc x + cot x) dx`

`=-log|cscx+ cotx| + c`

                            
slide-show version coming soon