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Question Number 10790 by Nur450737 last updated on 25/Feb/17

∫_0 ^(π/2) (√(sin x)) dx

$$\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \sqrt{\mathrm{sin}\:{x}}\:{dx} \\ $$

Answered by bahmanfeshki last updated on 26/Feb/17

(√(sin x))=t  t^2 =sin x⇒cos x=(√(1−t^2 ))  2t dt=cos x dx⇒dx=((2t)/(√(1−t^2 )))dt  ∫_0 ^1  ((2t^2 )/(√(1−t^2 )))dt=∫_0 ^(π/2) 2sin^2 vdv=[v−((sin 2v)/2)]_0 ^(π/2) =(π/2)

$$\sqrt{\mathrm{sin}\:{x}}={t} \\ $$$${t}^{\mathrm{2}} =\mathrm{sin}\:{x}\Rightarrow\mathrm{cos}\:{x}=\sqrt{\mathrm{1}−{t}^{\mathrm{2}} } \\ $$$$\mathrm{2}{t}\:{dt}=\mathrm{cos}\:{x}\:{dx}\Rightarrow{dx}=\frac{\mathrm{2}{t}}{\sqrt{\mathrm{1}−{t}^{\mathrm{2}} }}{dt} \\ $$$$\int_{\mathrm{0}} ^{\mathrm{1}} \:\frac{\mathrm{2}{t}^{\mathrm{2}} }{\sqrt{\mathrm{1}−{t}^{\mathrm{2}} }}{dt}=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \mathrm{2sin}\:^{\mathrm{2}} {vdv}=\left[{v}−\frac{\mathrm{sin}\:\mathrm{2}{v}}{\mathrm{2}}\right]_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} =\frac{\pi}{\mathrm{2}} \\ $$

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