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Hey,

(From an Integration Table)

Prove,

[tex]

{\int}{\sqrt {{{a}^{2}} - {{x}^{2}}}}{dx} = {{\frac {1}{2}}{\left(}{{{x}{\sqrt {{{a}^{2}} - {{x}^{2}}}}} + {{{a}^{2}}{\arcsin{\frac {x}{a}}}}}{\right)}}{,}{\,}{\,}{\,}{\,}{\,}{\,}{{{|}{x}{|}}{\leq}{{|}{a}{|}}}

[/tex]

Knowledge of Trigonometric Substitution for Integration ("Backwards" and "Forwards").

Knowledge of integration techniques involving the form,

[tex]

{\sqrt {{{a}^{2}} - {{x}^{2}}}}

[/tex]

Integration by Parts (IBP),

[tex]

{\int}{d{\left[}{u(x)}{v(x)}{\right]}} = {\int}{\biggl[}{u(x)}{d[v(x)]} + {v(x)}{d[u(x)]}{\biggl]}

[/tex]

[tex]

{\int}{u(x)}{d[v(x)]} = {{u(x)}{v(x)}} - {\int}{v(x)}{d[u(x)]}

[/tex]

What is bothering me about this integral is that I do not have a [itex]{x}[/itex] term on the outside of the radical which is preventing from evaluating this integral by normal convention.

Let,

[tex]

{I} = {\int}{\sqrt {{{a}^{2}} - {{x}^{2}}}}{dx}

[/tex]

In applying trigonometric substitution - consider the right triangle,

So,

[tex]

{{a}{\sin{\theta}}} = {\sqrt {{{a}^{2}} - {{x}^{2}}}}

[/tex]

Solving by "forward" substitution,

[tex]

{x} = {\pm}{\sqrt {{{a}^{2}}{{\cos}^{2}{\theta}}}}

[/tex]

However, how would you justify which root to take: the positive one or the negative one?

Thanks,

-PFStudent

## Homework Statement

(From an Integration Table)

Prove,

[tex]

{\int}{\sqrt {{{a}^{2}} - {{x}^{2}}}}{dx} = {{\frac {1}{2}}{\left(}{{{x}{\sqrt {{{a}^{2}} - {{x}^{2}}}}} + {{{a}^{2}}{\arcsin{\frac {x}{a}}}}}{\right)}}{,}{\,}{\,}{\,}{\,}{\,}{\,}{{{|}{x}{|}}{\leq}{{|}{a}{|}}}

[/tex]

## Homework Equations

Knowledge of Trigonometric Substitution for Integration ("Backwards" and "Forwards").

Knowledge of integration techniques involving the form,

[tex]

{\sqrt {{{a}^{2}} - {{x}^{2}}}}

[/tex]

Integration by Parts (IBP),

[tex]

{\int}{d{\left[}{u(x)}{v(x)}{\right]}} = {\int}{\biggl[}{u(x)}{d[v(x)]} + {v(x)}{d[u(x)]}{\biggl]}

[/tex]

[tex]

{\int}{u(x)}{d[v(x)]} = {{u(x)}{v(x)}} - {\int}{v(x)}{d[u(x)]}

[/tex]

## The Attempt at a Solution

What is bothering me about this integral is that I do not have a [itex]{x}[/itex] term on the outside of the radical which is preventing from evaluating this integral by normal convention.

Let,

[tex]

{I} = {\int}{\sqrt {{{a}^{2}} - {{x}^{2}}}}{dx}

[/tex]

In applying trigonometric substitution - consider the right triangle,

So,

[tex]

{{a}{\sin{\theta}}} = {\sqrt {{{a}^{2}} - {{x}^{2}}}}

[/tex]

Solving by "forward" substitution,

[tex]

{x} = {\pm}{\sqrt {{{a}^{2}}{{\cos}^{2}{\theta}}}}

[/tex]

However, how would you justify which root to take: the positive one or the negative one?

Thanks,

-PFStudent

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