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Standard Formulae - 1

Prove that: ...

Question

Prove that:
$\int \sqrt{a^{2} - x^{2}} d x = \frac{x}{2} \sqrt{a^{2} - x^{2}} + \frac{a^{2}}{2} {sin}^{- 1} (\frac{x}{a}) + c$

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Solution

$I = \int \sqrt{a^{2} - x^{2}} d x - - - - - (1)$
$I = [\sqrt{a^{2} - x^{2}} \int 1 - d x \int \frac{d}{d x} \sqrt{a^{2} - x^{2}} (\int 1 d x)] d x$
$I = x \sqrt{a^{2} - x^{2}} - \int \frac{- x^{2}}{\sqrt{a^{2} - x^{2}}} d x$
$I = x \sqrt{a^{2} - x^{2}} - \int \frac{- a^{2} + a^{2} - x^{2}}{\sqrt{a^{2} - x^{2}}} d x$
$I = x \sqrt{a^{2} - x^{2}} - \int \frac{- a^{2}}{\sqrt{a^{2} - x^{2}}} d x - \int \frac{a^{2} - x^{2}}{\sqrt{a^{2} - x^{2}}} d x$
$I = x \sqrt{a^{2} - x^{2}} + a^{2} \int \frac{1}{\sqrt{a^{2} - x^{2}}} d x - \int \sqrt{a^{2} - x^{2}} d x$
$I = x \sqrt{a^{2} - x^{2}} + a^{2} {sin}^{- 2} (\frac{x}{a}) - I + C$
$2 I = x \sqrt{a^{2} - x^{2}} + a^{2} {sin}^{- 2} (\frac{x}{a}) + C$
$I = \frac{x}{2} \sqrt{a^{2} - x^{2}} + \frac{a^{2}}{z} {sin}^{- 2} (\frac{x}{a}) + C$

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\int \sqrt{a^{2} - x^{2}} d x = \frac{x}{2} \sqrt{a^{2} - x^{2}} + \frac{a^{2}}{2} {sin}^{- 1} (\frac{x}{a}) + c

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\frac{d}{d x} (\frac{x}{2} \sqrt{a^{2} - x^{2}} + \frac{a^{2}}{2} {sin}^{- 1} \frac{x}{a}) = \sqrt{a^{2} - x^{2}}

\int \sqrt{x^{2} - a^{2}} d x = \frac{x}{2} \sqrt{x^{2} - a^{2}} - \frac{a^{2}}{2} log | x + \sqrt{x^{2} - a^{2}} | + c

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