If the value of the definite integral ∫01 sin−1⁡xx2−x+1dx is π2n (where n∈N)), then the value of n is __.

Solution:

$\begin{array}{l} I = \int_{0}^{1} \frac{\sin^{- 1} \sqrt{x}}{x^{2} - x + 1} d x \\ I = \int_{0}^{1} \frac{\sin^{- 1} \sqrt{1 - x}}{x^{2} - x + 1} d x = \int_{0}^{1} \frac{\cos^{- 1} \sqrt{x}}{x^{2} - x + 1} d x \end{array}$

On adding (1) and (2), we get

$\begin{array}{l} 2 I = \int_{0}^{1} \frac{\sin^{- 1} \sqrt{x} + \cos^{- 1} \sqrt{x}}{x^{2} - x + 1} d x \\ = \frac{π}{2} \int_{0}^{1} \frac{d x}{x^{2} - x + 1} d x \\ = \frac{π}{2} \int_{0}^{1} \frac{d x}{{(x - \frac{1}{2})}^{2} + {(\frac{\sqrt{3}}{2})}^{2}} d x \\ ∴ 2 I = \frac{π}{2} \frac{1}{(\frac{\sqrt{3}}{2})} {[\tan^{- 1} (\frac{2 x - 1}{\sqrt{3}})]}_{0}^{1} = \frac{π^{2}}{3 \sqrt{3}} \end{array}$

Hence, $I = \frac{π^{2}}{6 \sqrt{3}} = \frac{π^{2}}{\sqrt{108}} \equiv \frac{π^{2}}{\sqrt{n}}$

If the value of the definite integral $\int_{0}^{1} \frac{\sin^{- 1} \sqrt{x}}{x^{2} - x + 1} d x$ is $\frac{π^{2}}{\sqrt{n}}$ (where $n \in N)$ ), then the value of n is __.

Solution:

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