What is the Integral of a Trigonometric Function with a Weird Substitution?

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SUMMARY

The integral of the function \(\int_{0}^{\frac{\pi}{2}}\frac{\sqrt{\sin x}}{\sqrt{\sin x}+\sqrt{\cos x}}\mathrm{d}x\) can be effectively solved using the substitution \(y=\frac{\pi}{2}-x\). This substitution leverages the trigonometric identities \(\sin(\frac{\pi}{2} - x) = \cos x\) and \(\cos(\frac{\pi}{2} - x) = \sin x\) to simplify the integral. The discussion highlights the importance of correctly applying these identities to achieve a solution.

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Homework Statement



I need to calculate the integral:
[tex]\int_{0}^{\frac{\pi}{2}}\frac{\sqrt{\sin x}}{\sqrt{\sin x}+\sqrt{\cos x}}\mathrm{d}x[/tex]

Homework Equations



The Attempt at a Solution


There is a tip: "try substituting [tex]y=\frac{\pi}{2}-x[/tex]. I tried it and didn't get anywhere. I also tried several trigonometric identities.

Thanks in advance!
 
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Yeah you should probably try the suggested substitution again. Are you using the fact that sin(pi/2 - x) = cos x and cos(pi/2 - x) = sin x?
 
Ok, I got it now... Thanks! :)
 

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