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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Question statement

Find the following integral.

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			Name	Type	Generated Value

a	integer	8
c	integer	5
b	integer	9
b1	integer	8

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$I=\simplify[std]{Int(({c} / (sqrt({a}-{b}x^2))),x)}$

$I=\;$ Unnamed gap

Input all numbers as integers, fractions or surds. No decimal numbers. You input surds, for example $\sqrt{2}$ , by writing sqrt(2).

Input the constant of integration as $C$ .

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Advice

For the integral

$I=\simplify[std]{Int((({c}) / (sqrt({a}-{b}x^2))),x)}$ use the substitution

$\displaystyle \simplify[std]{u=(sqrt({b})/sqrt({a}))*x}$
so that

$\simplify[all,!sqrtProduct,fractionNumbers]{sqrt({a}-{b}x^2)=sqrt({a}-{b}*({a}/{b})*u^2)=sqrt({a}-{a}*u^2)=sqrt({a})*sqrt(1-u^2)}$

We have $\displaystyle \simplify[std]{du=(sqrt({b})/sqrt({a}))dx}$ and we get

$\begin{eqnarray*}I&=&\simplify[std]{({c}*(sqrt({a})/sqrt({b})))*Int((1 / ( sqrt({a})*sqrt(1-u^2) )),u)}\\ &=&\simplify[std]{({c}/sqrt({b}))*Int((1 / (sqrt(1-u^2))),u)}\\ &=&\simplify[std]{({c}/sqrt({b}))*arcsin(u)+C}\\ &=&\simplify[std]{({c}/sqrt({b}))*arcsin((sqrt({b})/sqrt({a}))*x)+C} \end{eqnarray*}$
on replacing

$u$ by

$\displaystyle \simplify[std]{(sqrt({b})/sqrt({a}))*x}$

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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`