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

                      I=∫_0 ^1 ((sin^(−1) (√x))/( (√(1−x+x^2 ))))dx=(π/4)ln3

$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:{I}=\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\mathrm{sin}^{−\mathrm{1}} \sqrt{{x}}}{\:\sqrt{\mathrm{1}−{x}+{x}^{\mathrm{2}} }}{dx}=\frac{\pi}{\mathrm{4}}{ln}\mathrm{3} \\ $$

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

I=∫_0 ^1 ((sin^(−1) (√x))/( (√(1−x+x^2 ))))dx(1)=∫_0 ^1 ((sin^(−1) (√(1−x)))/( (√(x+(1−x)^2 ))))dx=∫_0 ^1 ((sin^(−1) (√(1−x)))/( (√(1−x+x^2 ))))dx(2)  (1)+(2)=2I=(π/2)∫_0 ^1 (dx/( (√(1−x+x^2 ))))=(π/2)ln3  ⇒I=(π/4)ln3    ...... ✓✓......

$${I}=\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\mathrm{sin}^{−\mathrm{1}} \sqrt{{x}}}{\:\sqrt{\mathrm{1}−{x}+{x}^{\mathrm{2}} }}{dx}\left(\mathrm{1}\right)=\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\mathrm{sin}^{−\mathrm{1}} \sqrt{\mathrm{1}−{x}}}{\:\sqrt{{x}+\left(\mathrm{1}−{x}\right)^{\mathrm{2}} }}{dx}=\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\mathrm{sin}^{−\mathrm{1}} \sqrt{\mathrm{1}−{x}}}{\:\sqrt{\mathrm{1}−{x}+{x}^{\mathrm{2}} }}{dx}\left(\mathrm{2}\right) \\ $$$$\left(\mathrm{1}\right)+\left(\mathrm{2}\right)=\mathrm{2}{I}=\frac{\pi}{\mathrm{2}}\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{dx}}{\:\sqrt{\mathrm{1}−{x}+{x}^{\mathrm{2}} }}=\frac{\pi}{\mathrm{2}}{ln}\mathrm{3} \\ $$$$\Rightarrow{I}=\frac{\pi}{\mathrm{4}}{ln}\mathrm{3}\:\:\:\:......\:\checkmark\checkmark...... \\ $$

Commented by mnjuly1970 last updated on 24/Mar/21

    very nice thanks alot...

$$\:\:\:\:{very}\:{nice}\:{thanks}\:{alot}... \\ $$

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