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Solution suggested by @John Muradeli

$\begin{aligned} I & = \int_0^1 \frac {x\color{#3D99F6}\tan \left(\cos^{-1}x\right)}{\sqrt{1-x^2}} dx & \small \color{#3D99F6} \text{See note.} \\ & = \int_0^1 \frac x{\sqrt{1-x^2}} \cdot {\color{#3D99F6}\frac {\sqrt{1-x^2}}{x}} dx \\ & = \int_0^1 dx = \boxed{1} \end{aligned}$

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Note: Let $\color{#3D99F6} \cos^{-1} x = \theta$ , then $\tan \left({\color{#3D99F6}\cos^{-1} x} \right) = \tan {\color{#3D99F6} \theta} = \dfrac {\sin \theta}{\cos \theta} = \dfrac {\sqrt{1-x^2}}{x}$

$I=\int _{ 0 }^{ 1 }{ \frac { tan({ cos }^{ -1 }x) }{ \sqrt { 1-{ x }^{ 2 } } } xdx }$

Putting $x=cos(\theta)$ we have :

$\Rightarrow \theta ={ cos }^{ -1 }x$

$\Rightarrow \frac { -dx }{ \sqrt { 1-{ x }^{ 2 } } } =d\theta$

Evaluating the limits we have limits as $x=0 , \theta=\frac{\pi}{2}$ and $x=1, \theta = 0$

$I=-\int _{ \frac { \pi }{ 2 } }^{ 0 }{ tan(\theta) cos(\theta) d\theta }$

$I=\int _{ 0 }^{ \frac { \pi }{ 2 } }{ tan(\theta) cos(\theta) d\theta } =\int _{ 0 }^{ \frac { \pi }{ 2 } }{ sin(\theta) d\theta } =1$

Hint:

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

is an identity.

You can also substitute $\cos^{-1}x = u$ . Then $\frac {dx}{\sqrt {1 - x^{2}}} = -du$ . The integral becomes -

$\int_{\frac {1}{2}\pi}^{\frac {1}{4}\pi} \cos u \cdot \tan u (-du) = \int_{\frac {1}{4}\pi}^{\frac {1}{2}\pi} \sin u du = -\cos u \Big|_{\frac {1}{4}\pi}^{\frac {1}{2}\pi} = \boxed {1}$