Question

Hello any genius here to solve

Prove that
integration of 1/a² - x² = sin^(-1) (x/a) + C

Thank you! ​

Answer 1

Answer:

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Step-by-step explanation:

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Answer 2

Step-by-step explanation:

$\huge\blue{\underline{\underline{\bf Question:-}}}$

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

$\huge\blue{\underline{\underline{\bf Answer:-}}}$

$\implies \int \frac{1}{ {a}^{2} - {x}^{2} } \\ \\ let \: \: x = a\sin( \theta) \\ \\ \frac{dx}{d \theta} = a \cos( \theta) \\ \\ dx = a \cos( \theta) \: d( \theta) \\ \\ \implies \int \frac{1}{ {a}^{2} - {x}^{2} } \\ \\ \implies \int \frac{1}{ {a}^{2} - {a}^{2} { \sin}^{2}( \theta) } \times a\cos( \theta) \: d( \theta) \\ \\ \implies \int \frac{ \cancel{a}\cos( \theta) \: d( \theta) }{ \cancel{a} \sqrt{1 - { \sin}^{2} ( \theta)} } \\ \\ \implies \int \frac{ \cancel { \cos( \theta)}d( \theta)}{ \cancel{\cos( \theta)} } \\ \\ \implies \int d( \theta) \\ \\ \implies ( \theta) + c .....(1)\\ \\ x = a \sin( \theta) \\ ( \theta) = { \sin}^{ - 1} ( \frac{x}{a} ) \\ \\ putting \: \: value \: \: of \: \: ( \theta) \: \: in \: \: (1) \\ \\ \implies ( \theta) + c \\ \\ \huge \boxed{\leadsto ( { \sin}^{ - 1} ( \frac{x}{a} ) + c)}$