Indefinite integral by substitution

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Bill Foster

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Bill Foster 55 years, 3 months ago

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Name	Status	Author	Last Modified
Indefinite integral by substitution	draft	Bill Foster	06/08/2020 08:41
Indefinite integral by substitution	draft	Julie Crowley	08/10/2016 21:58

Question statement

Find the following integral.

Input the constant of integration as $C$ .

Input all numbers as integers or fractions not as decimals.

Give any introductory information the student needs.

			Name	Type	Generated Value

a	integer	4
s1	integer	1
b	integer	5

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xxxxxxxxxx
 
2*s1*random(1..5)

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Generated value: `integer`

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$I=\int\frac{\simplify[std]{{a}*x+{b}}}{(1-x^2)^{1/2}} \;dx$

$I=\;$ Unnamed gap

Input the constant of integration as $C$ .

Input all numbers as integers or fractions not as decimals.

Click on Show steps if you need help. You will lose 1 mark if you do so.

Advice

Split the integral into two parts

$I=\int\frac{\simplify[std]{{a}*x}}{(1-x^2)^{1/2}} \;dx+\int\frac{\simplify[std]{{b}}}{(1-x^2)^{1/2}} \;dx$
For the integral

$I_1=\int\frac{\simplify[std]{{a}*x}}{(1-x^2)^{1/2}} \;dx$ use the substitution

$u=1-x^2$ and then

$du=-2xdx$ and we get

$\begin{eqnarray*}I_1&=&\simplify[std]{{-a}/2*Int((1 / (u^(1/2))),u)}\\\\ &=&\simplify[std]{({-a}/2)*(2u^(1/2))+C}\\ &=&\simplify[std]{({-a})*(1-x^2)^(1/2)+C} \end{eqnarray*}$
The other integral is a standard result:

$I_2=\simplify[std]{Int((({b}) / (1-x^2)^(1/2)),x)={b}*arcsin(x)+C}$
Putting these together gives:

$I=I_1+I_2=\simplify[std]{-{a}*(1-x^2)^(1/2)+{b}*arcsin(x)+C}$

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Taxonomy: mathcentre

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Bill Foster 55 years, 3 months ago

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Generated value: `integer`

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Indefinite integral by substitution

Metadata

Taxonomy: mathcentre

Taxonomy: Kind of activity

Taxonomy: Context

Contributors

History

Comment

Checkpoint description

Bill Foster 55 years, 3 months ago

Error in variable testing condition

Error

Warning

Generated value: integer

Parts

Gap-fill

Generated value: `integer`