∫_0 ^( (1/2)) (√(1+(√(1−x^2 )))) dx =?


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Question Number 133173 by john_santu last updated on 19/Feb/21

$\int_{0}^{\frac{1}{2}} \sqrt{1 + \sqrt{1 - x^{2}}} dx = ?$
	Answered by EDWIN88 last updated on 19/Feb/21

	$I = \int_{0}^{\frac{1}{2}} \sqrt{1 + \sqrt{1 - x^{2}}} dx$ $by substitution \sqrt{1 + \sqrt{1 - x^{2}}} = \sqrt{2 - s^{2}}$ $\Rightarrow \sqrt{1 - x^{2}} = 1 - s^{2}; x = s \sqrt{2 - s^{2}}$ $\frac{dx}{ds} = \sqrt{2 - s^{2}} - \frac{s^{2}}{\sqrt{2 - s^{2}}} = \frac{2 - {2 s}^{2}}{\sqrt{2 - s^{2}}}$ $I = \int_{0}^{\frac{\sqrt{3} - 1}{2}} (2 - {2 s}^{2}) ds = {[(2 s - \frac{2}{3} s^{3})]}_{0}^{\frac{\sqrt{3} - 1}{2}}$ $= \frac{3 \sqrt{3} - 1}{6} .$
	Answered by Dwaipayan Shikari last updated on 19/Feb/21

	$\sqrt{1 + \sqrt{1 - x^{2}}} = \sqrt{\frac{1 + x}{2}} + \sqrt{\frac{1 - x}{2}}$ $= \frac{1}{\sqrt{2}} \int_{0}^{\frac{1}{2}} \sqrt{1 + x} + \frac{1}{\sqrt{2}} \int_{0}^{\frac{1}{2}} \sqrt{1 - x} d x$ $= \frac{\sqrt{2}}{3} {(\frac{3}{2})}^{\frac{3}{2}} - \frac{\sqrt{2}}{3} - \frac{\sqrt{2}}{3} {(\frac{1}{2})}^{\frac{3}{2}} + \frac{\sqrt{2}}{3} = \frac{3 \sqrt{3}}{6} - \frac{1}{6}$
	Answered by mathmax by abdo last updated on 20/Feb/21
	$I=∫_0 ^(1/2) (√(1+(√(1−x^2 ))))dx we do the changement x=sint ⇒ I =∫_0 ^(π/6) (√(1+cost))cost dt =∫_0 ^(π/6) (√2)cos((t/2))cost dt =(√2)∫_0 ^(π/6) (1/2){cos(t+(t/2))+cos(t−(t/2))}dt =((√2)/2)∫_0 ^(π/6) cos(((3t)/2))dt +((√2)/2)∫_0 ^(π/6) cos((t/2))dt =((√2)/2)×(2/3)[sin(((3t)/2))]_0 ^(π/6) +((√2)/2)×2[sin((t/2))]_0 ^(π/6) =((√2)/3)×sin((3/2).(π/6)) +(√2)sin((π/(12))) =((√2)/3)sin((π/4))+(√2)sin((π/(12))) =((√2)/3).(1/( (√2))) +(√2).((√(2−(√3)))/2) =(1/3)+((√(2−(√3)))/( (√2)))=(1/3)+(√(1−((√3)/2)))$
	$I = \int_{0}^{\frac{1}{2}} \sqrt{1 + \sqrt{1 - x^{2}}} dx we do the changement x = sint \Rightarrow$ $I = \int_{0}^{\frac{π}{6}} \sqrt{1 + cost} cost dt = \int_{0}^{\frac{π}{6}} \sqrt{2} \cos (\frac{t}{2}) cost dt$ $= \sqrt{2} \int_{0}^{\frac{π}{6}} \frac{1}{2} {\cos (t + \frac{t}{2}) + \cos (t - \frac{t}{2})} dt$ $= \frac{\sqrt{2}}{2} \int_{0}^{\frac{π}{6}} \cos (\frac{3 t}{2}) dt + \frac{\sqrt{2}}{2} \int_{0}^{\frac{π}{6}} \cos (\frac{t}{2}) dt$ $= \frac{\sqrt{2}}{2} \times \frac{2}{3} {[\sin (\frac{3 t}{2})]}_{0}^{\frac{π}{6}} + \frac{\sqrt{2}}{2} \times 2 {[\sin (\frac{t}{2})]}_{0}^{\frac{π}{6}}$ $= \frac{\sqrt{2}}{3} \times \sin (\frac{3}{2} . \frac{π}{6}) + \sqrt{2} \sin (\frac{π}{12})$ $= \frac{\sqrt{2}}{3} \sin (\frac{π}{4}) + \sqrt{2} \sin (\frac{π}{12}) = \frac{\sqrt{2}}{3} . \frac{1}{\sqrt{2}} + \sqrt{2} . \frac{\sqrt{2 - \sqrt{3}}}{2}$ $= \frac{1}{3} + \frac{\sqrt{2 - \sqrt{3}}}{\sqrt{2}} = \frac{1}{3} + \sqrt{1 - \frac{\sqrt{3}}{2}}$
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