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No mathematician in its right mind will tackle this integral head-on. So, we will tackle this integral butt-on.

Take its limbs out: First of all, notice that each of the groups of expressions on top match in some nature to the expressions on the bottom. The first easily distinguishable match is the combo of $a$ expressions:

$\frac{(\frac { a+x }{ \sqrt [ 3 ]{ a^{ 2 } } -\sqrt [ 3 ]{ x^{ 2 } } } -\sqrt[6]{x}(\sqrt[6]{a}-\sqrt[6]{x})+\frac { \sqrt [ 3 ]{ ax^{ 2 } } -\sqrt [ 3 ]{ a^{ 2 }x } }{ \sqrt [ 3 ]{ a^{ 2 } } -2\sqrt [ 3 ]{ ax } +\sqrt [ 3 ]{ x^{ 2 } } })}{(\sqrt[6]a-\sqrt[6]{x})}$

These are the legs. Now let's make a soup, shall we?: First of all, notice that $\sqrt[6]{x}$ is an ulcer; that's not tasty.

$\frac{(\frac { a+x }{ \sqrt [ 3 ]{ a^{ 2 } } -\sqrt [ 3 ]{ x^{ 2 } } } +\frac { \sqrt [ 3 ]{ ax^{ 2 } } -\sqrt [ 3 ]{ a^{ 2 }x } }{ \sqrt [ 3 ]{ a^{ 2 } } -2\sqrt [ 3 ]{ ax } +\sqrt [ 3 ]{ x^{ 2 } } })}{(\sqrt[6]a-\sqrt[6]{x})}-\sqrt[6]{x}.$

Way better. Ok, let's strip the bones:

$\frac { a+x }{ \sqrt [ 3 ]{ a^{ 2 } } -\sqrt [ 3 ]{ x^{ 2 } } } -\frac { \sqrt [ 3 ]{ ax^{ 2 } } -\sqrt [ 3 ]{ a^{ 2 }x } }{ \sqrt [ 3 ]{ a^{ 2 } } -2\sqrt [ 3 ]{ ax } +\sqrt [ 3 ]{ x^{ 2 } } }$

$\Rightarrow \frac { (\sqrt [ 3 ]{ a } )^{ 3 }+(\sqrt [ 3 ]{ x } )^{ 3 } }{ (\sqrt [ 3 ]{ a } )^{ 2 }-(\sqrt [ 3 ]{ x } )^{ 2 } } -\frac { \sqrt [ 3 ]{ ax } (\sqrt [ 3 ]{ a } -\sqrt [ 3 ]{ x } ) }{ (\sqrt [ 3 ]{ a } -\sqrt [ 3 ]{ x } )^{ 2 } }$ (Let $a=\sqrt[3]{a}$ and $b=\sqrt[3]{x}$ )

$\Rightarrow \frac { (a+b)(a^{ 2 }-ab-b^{ 2 }) }{ a^{ 2 }-b^{ 2 } } -\frac { ab }{ a-b }$

$\Rightarrow \frac { (a+b)(a^{ 2 }-ab-b^{ 2 }) }{ a^{ 2 }-b^{ 2 } } \cdot \frac { a-b }{ a-b } -\frac { ab }{ a-b } \cdot \frac { a^{ 2 }-b^{ 2 } }{ a^{ 2 }-b^{ 2 } }$

$\Rightarrow \frac { (a^{ 2 }-b^{ 2 })(a^{ 2 }-ab-b^{ 2 }) }{ (a^{ 2 }-b^{ 2 })(a-b) } -\frac { (a^{ 2 }-b^{ 2 })ab }{ (a^{ 2 }-b^{ 2 })(a-b) }$

$\Rightarrow \frac { (a^{ 2 }-b^{ 2 })(a^{ 2 }-ab-b^{ 2 }) }{ (a^{ 2 }-b^{ 2 })(a-b) } -\frac { (a^{ 2 }-b^{ 2 })ab }{ (a^{ 2 }-b^{ 2 })(a-b) }$

$\Rightarrow \frac { (a^{ 2 }-b^{ 2 })(a^{ 2 }-2ab-b^{ 2 }) }{ (a^{ 2 }-b^{ 2 })(a-b) }$

$\Rightarrow \frac{(a^2-b^2)(a-b)^2}{(a^2-b^2)(a-b)}$

$\Rightarrow a-b$

$\Rightarrow \sqrt[3]{a}-\sqrt[3]{x}$

Now insert back into the original expression:

$\frac{\sqrt[3]{a}-\sqrt[3]{x}}{\sqrt[6]{a}-\sqrt[6]{x}}-\sqrt[6]{x}$

$\Rightarrow \frac{a^2-b^2}{a-b}-b$

$\Rightarrow \frac{(a-b)(a+b)}{a-b}-b$

$\Rightarrow a+b-b$

$\Rightarrow \sqrt[6]{a}$

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Ok, the legs are steaming. Now our integral looks like this:

$\int _{ 0 }^{ a^{ -\frac { 1 }{ 6 } } }{\sqrt[6]{a} \frac { \tan { (\cos ^{ -1 }{ x } ) } (\sqrt { \sqrt { 2 } -1 } \sqrt [ 4 ]{ 3+2\sqrt { 2 } } +\sqrt [ 3 ]{ (x+12)\sqrt { 2 } -6x-8 } )x }{ (\frac { x-\sqrt { x } }{ \sqrt { x } -1 } -\sqrt { \sqrt { 2 } +1 } \sqrt [ 4 ]{ 3-2\sqrt { 2 } } )\sqrt { 1-x^{ 2 } } } dx }.$

Still fighting back. Ok, let's take its arms off: notice now we have an awkward expression on top and on bottom; let's take it out:

$\frac{(\sqrt { \sqrt { 2 } -1 } \sqrt [ 4 ]{ 3+2\sqrt { 2 } } +\sqrt [ 3 ]{ (x+12)\sqrt { 2 } -6x-8 } )}{(\frac { x-\sqrt { x } }{ \sqrt { x } -1 } -\sqrt { \sqrt { 2 } +1 } \sqrt [ 4 ]{ 3-2\sqrt { 2 } } )}$ .

Now, to cook the arms:

$\sqrt{\sqrt{2}-1}\sqrt[4]{3+2\sqrt{2}}=\sqrt[4]{(\sqrt{2}-1)^2(3+2\sqrt{2})}=1.$

An analogous simplification may be applied in the denominator. The variable expression in the numerator:

$\sqrt[3]{(x+12)\sqrt{x}-6x-8}$ $\Rightarrow \sqrt[3]{x\sqrt{x}+12\sqrt{x}-6x-8}$

$\Rightarrow \sqrt[3]{g^3-6g^2+12g-8}$

$\Rightarrow \sqrt[3]{(g-2)^3}$ $\Rightarrow \sqrt[3]{(\sqrt{x}-2)^3}$

$\Rightarrow \sqrt{x}-2$ .

The fraction $\frac{x-\sqrt{x}}{\sqrt{x}-1}$ simplifies to: $\frac{\sqrt{x}(\sqrt{x}-1)}{\sqrt{x}-1}=\sqrt{x}$

And thus, $\frac{1+\sqrt{x}-2}{\sqrt{x}-1}=\frac{\sqrt{x}-1}{\sqrt{x}-1}=1$ .

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No arms no legs; now it's face to face: $\int _{ 0 }^{ a^{ -\frac { 1 }{ 6 } } }{ \frac { \tan { (\cos ^{ -1 }{ x } ) } x }{ \sqrt { 1-x^{ 2 } } } dx }$

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Take its head off. Rewrite as follows: $\tan { (\cos ^{ -1 }{ x } ) } \cdot \frac{x}{{ \sqrt { 1-x^{ 2 } } }}$

Notice a familiarity between the fraction and the arctangent. In fact, it's an inobvious identity:

$\tan {(\cos^{-1}{x})}=\frac{\sqrt{1-x^2}}{x}$ .

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Proof: Let angle $\varphi =\cos^{-1}{x}$ be bound between 0° and 180° (by the definition of the principal value of arccosine). We have $\cos{\varphi}=x$ . Then,

sssss

$\sin{\varphi}=+\sqrt{1-x^2}$ .

Thus,

$\tan{\varphi}=\frac{\sqrt{1-x^2}}{x}$ ,

or,

$\tan{(\cos^{-1}{x})}=\frac{\sqrt{1-x^2}}{x}$ .

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And so, we have 1 (as long as $x\neq 0$ , which implies that $a\neq 0$ , which satisfies the initial condition).

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And now, let's put it all together into a one big dish with raw torso:

$\int _{0 }^{a^{-\frac{1}{6}} }{1\cdot 1\cdot \sqrt[6]{a}dx } =\int _{0 }^{a^{-\frac{1}{6}} }{ \sqrt[6]{a}dx }$

And finally, IT'S DENTAL CLUBBERING TIME!!!

$\int _{0 }^{a^{-\frac{1}{6}} }{ \sqrt[6]{a}dx }$

$={\left[ \sqrt[6]{a}x \right]}_{0}^{a^{-\frac{1}{6}}}$

$=\sqrt[6]{a}\cdot a^{-\frac{1}{6}}-\sqrt[6]{a}\cdot 0=\sqrt[6]{a}\cdot \frac{1}{\sqrt[6]{a}}$

$=\boxed{1}$

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Yhum...

Oh COME on!