Question

Integral of 1 over a square minus x square square root

Answer 1

$: \implies \tt \:Evaluate \: \int\dfrac{dx}{ \sqrt{ {a}^{2} - {x}^{2} } }$

$\bf \:\large \red{AηsωeR }$

$: \implies \tt \:Let \: I \: = \int\dfrac{dx}{ \sqrt{ {a}^{2} - {x}^{2} } }$

$: \implies \red{ \sf \: Put \: x \: = \: a \: sin \theta \: } - - (1)$

● On differentiating both sides, we get

$: \implies \tt \: \red{ \sf \: \dfrac{dx}{d\theta} = a \: cos\theta}$

$: \implies \red{ \bf \: dx \: = \: a \: cos\theta \: d\theta}$

● On substituting these values in given integral, we get

$: \implies \tt \: I = \int\dfrac{a \: cos\theta \: d\theta}{ \sqrt{ {a}^{2} - {a}^{2} {sin}^{2}\theta } }$

$: \implies \tt \: I = \int \: \dfrac{a \: cos\theta \: d\theta}{ \sqrt{ {a}^{2}(1 - {sin}^{2} \theta)} }$

$: \implies \tt \: I \: = \int \: \dfrac{a \: cos\theta \: d\theta}{ \sqrt{ {a}^{2} \: {cos}^{2} \theta } }$

$: \implies \tt \: I \: = \int \: \dfrac{a \: cos\theta \: d\theta}{a \: cos\theta}$

$: \implies \tt \: I \: = \int \: d\theta \:$

$: \implies \tt \: I \: = \theta \: + \: c$

$: \implies \tt \: I \: = \: {sin}^{ - 1} \bigg( \dfrac{x}{a} \bigg) + \: c$

$\bigg( \because \tt\: from \: (1) \: sin\theta = \dfrac{x}{a} \bf\implies \:\theta = {sin}^{ - 1} \dfrac{x}{a} \bigg)$

$\large{\boxed{\boxed{\bf{Hence, \int\dfrac{dx}{ \sqrt{ {a}^{2} - {x}^{2} } } = \: {sin}^{ - 1} \bigg( \dfrac{x}{a} \bigg) + \: c}}}}$