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mathsIntegral CalculusVarious Forms and Results of Integrals

### Integration by Substitution

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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

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