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Various Forms and Results of Integrals

Various Forms and Results of Integrals

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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`

Integration by Substitution

plain and simple summary

nub

plain and simple summary

nub

dummy

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

simple steps to build the foundation

trek

simple steps to build the foundation

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


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

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
  • 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".

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`
  • 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.

comprehensive information for quick review

Jogger

comprehensive information for quick review

Jogger

dummy

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`



           

practice questions to master the knowledge

Exercise

practice questions to master the knowledge

Exercise

Integrate `int tan x dx`

  • 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`

Integrate `int cot x dx`

  • 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`

Integrate `int sec x dx`

  • 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 `

Integrate `int csc x dx`

  • 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`

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Progress

Consider the integration integral 2x, sin x squared, dx. We notice that derivative of x squared, 2x, is part of the integration. Which of the following can be used ot work out the integration.
take;t;tea;substitute;integrand
take t = x squared and substitute in the integrand
noticing;does;not;help
noticing derivative of x squared; 2x does not help
The answer is "take t = x squared and substitute in the integrand"
Integration of 2x sine x^2 d x is given.
Considering the integration integral f of x d x: If it is observed that f of x can be rewritten as f of x =, h of g of x, multiplied g prime of x then, integral f of x d x, = ,int h of y d y. Where y = g of x and d y = g prime of x d x. This method is integration by substitution.
Simplified nub for integration by substitution is given
Reference Jogger for Integration by Substitution is given.
Given a function in parametric form y= r cos t and x=r sine t , how to proceed with the integration, integral y dx.
substitute;y;t;proceed
substitute y = r cos t and dx = derivative of r cos t. Then proceed with variable of integration as t
function;parametric;cannot;not;form
the function in parametric form cannot be integrated
The answer is "substitute y = r cos t and dx = derivative of r cos t"
Given function in parametric form, approach to integration is given.
Integrate: integral tan x dx.
1
2
The answer is "can be integrated", as given.
integrate cot x
1
2
The answer is "can be integrated", as given
Integrate secant x
1
2
The answer is "can be integrated", as given
Integrate co-secant x
1
2
The answer is "can be integrated", as given

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