Question

Integrate:–
$\Huge{{\large{∫} \frac{dx}{ \sqrt{x \sqrt{x} - x {}^{2} } }}}$
​

Answer 1

$\large\underline{\sf{Solution-}}$

$\rm :\longmapsto\:\displaystyle\int \: \dfrac{dx}{ \sqrt{x \sqrt{x} - {x}^{2} } }$

$\rm \: \: = \: \:\displaystyle\int \: \dfrac{dx}{ \sqrt{x \sqrt{x} - x \sqrt{x} \sqrt{x} } }$

$\rm \: \: = \: \:\displaystyle\int \:\dfrac{dx}{ \sqrt{x \sqrt{x}(1 - \sqrt{x} )} }$

$\rm \: \: = \: \:\displaystyle\int \:\dfrac{dx}{ \sqrt{x \sqrt{x}} \sqrt{1 - \sqrt{x} } }$

$\rm \: \: = \: \:\displaystyle\int \:\dfrac{dx}{ { \bigg(x \bigg)}^{ \dfrac{3}{4} } \sqrt{1 - \sqrt{x} } }$

$\red{\rm :\longmapsto\:Put \: \sqrt{1 - \sqrt{x} } = y}$

$\red{\rm :\longmapsto\:1 - \sqrt{x} = {y}^{2}}$

$\red{\rm :\longmapsto\:1 - {y}^{2} =\sqrt{x}}$

$\red{\rm :\longmapsto\:x = {(1 - {y}^{2} )}^{2}}$

$\red{\rm :\longmapsto\:dx = 2{(1 - {y}^{2} )}( - 2y) \: dy}$

$\red{\rm :\longmapsto\:dx = - 4y{(1 - {y}^{2} )} \: dy}$

So, on substituting all these values, we get

$\rm \: \: = \: \:\displaystyle\int \:\dfrac{ - 4 \: \cancel y \: (1 - {y}^{2}) \: dy }{ {\bigg( {(1 - {y}^{2}) }^{2} \bigg) }^{\dfrac{3}{4} } \times \cancel y}$

$\rm \: \: = \: \:\displaystyle\int \:\dfrac{ - 4(1 - {y}^{2}) \: dy }{ {\bigg( {1 - {y}^{2}}\bigg) }^{\dfrac{3}{2} }}$

$\rm \: \: = 4 \: \:\displaystyle\int \:\dfrac{ - 1}{ \sqrt{1 - {y}^{2} } } \: dy$

$\rm \: \: = \: \: 4 \: {cos}^{ - 1} y \: + \: c$

$\rm \: \: = \: \: 4 \: {cos}^{ - 1} \sqrt{1 - \sqrt{x} } \: + \: c$

Hence,

$\rm :\longmapsto\:\displaystyle\int \: \dfrac{dx}{ \sqrt{x \sqrt{x} - {x}^{2} } } \rm =4 \: {cos}^{ - 1} \sqrt{1 - \sqrt{x} } \: + \: c$

Additional Information :-

$\rm :\longmapsto\:\displaystyle\int\dfrac{dx}{ {x}^{2} + {a}^{2} } = \dfrac{1}{a} {tan}^{ - 1}\dfrac{x}{a} + c$

$\rm :\longmapsto\:\displaystyle\int\dfrac{dx}{ {x}^{2} - {a}^{2} } = \dfrac{1}{2a} \: {log} \: \dfrac{x - a}{x + a} \: + \: c$

$\rm :\longmapsto\:\displaystyle\int \: \dfrac{dx}{ \sqrt{ {a}^{2} - {x}^{2}}} = {sin}^{ - 1}\dfrac{x}{a} + c$

$\rm :\longmapsto\:\displaystyle\int \: \dfrac{dx}{ \sqrt{ {x}^{2} - {a}^{2}}} = log | \: x + \sqrt{ {x}^{2} - {a}^{2} } \: | + c$

$\rm :\longmapsto\:\displaystyle\int \: \dfrac{dx}{ \sqrt{ {x}^{2} + {a}^{2}}} = log | \: x + \sqrt{ {x}^{2} + {a}^{2} } \: | + c$