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Thought-Process to Discover Knowledge

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summary of this topic

Various Forms and Results of Integrals

Voice

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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

trek

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

Keep tapping on the content to continue learning.
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

Progress

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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