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Question Number 136643 by Ñï= last updated on 24/Mar/21

$$I={\int }_0^1\frac{{\sin }^{-1}\sqrt{x}}{\sqrt{1-x+x^2}}dx=\frac{\pi }{4}ln3$$

Answered by Ñï= last updated on 24/Mar/21

$$I={\int }_0^1\frac{{\sin }^{-1}\sqrt{x}}{\sqrt{1-x+x^2}}dx\left(1\right)={\int }_0^1\frac{{\sin }^{-1}\sqrt{1-x}}{\sqrt{x+{(1-x)}^2}}dx={\int }_0^1\frac{{\sin }^{-1}\sqrt{1-x}}{\sqrt{1-x+x^2}}dx\left(2\right)$$$$\left(1\right)+\left(2\right)=2I=\frac{\pi }{2}{\int }_0^1\frac{dx}{\sqrt{1-x+x^2}}=\frac{\pi }{2}ln3$$$$\Rightarrow I=\frac{\pi }{4}ln3......✓✓......$$

Commented by mnjuly1970 last updated on 24/Mar/21

$$verynicethanksalot...$$

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