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I have to calculate the integral:

$ \intop_{0}^{\frac{\pi}{2}}\frac{\sqrt{\sin x}}{\sqrt{\sin x}+\sqrt{\cos x}}dx$

I tried a various sort of ways, I'll present 2 of them that lead me nowhere, maybe someone will see a way through the obstacles.

way 1: trigonometric substitution :

substitue:

$ \tan\left(\frac{x}{2}\right)=t $

thus

$ \sin\left(x\right)=\frac{2t}{1+t^{2}},\cos\left(x\right)=\frac{1-t^{2}}{1+t^{2}},dx=\frac{2}{1+t^{2}}dt $

$ \intop_{0}^{\frac{\pi}{2}}\frac{\sqrt{\sin x}}{\sqrt{\sin x}+\sqrt{\cos x}}dx=\intop_{0}^{1}\frac{\frac{\sqrt{2t}}{\sqrt{1+t^{2}}}}{\frac{\sqrt{2t}}{\sqrt{1+t^{2}}}+\frac{\sqrt{1-t^{2}}}{\sqrt{1+t^{2}}}}\frac{2}{1+t^{2}}dt=\intop_{0}^{1}\frac{2\sqrt{2t}}{\left(1+t^{2}\right)\left(\sqrt{2t}+\sqrt{1-t^{2}}\right)}dt $

From here I cannot see how to continue.( I tried to multiply the denominator and the numerator by $ \sqrt{2t}-\sqrt{1-t^{2}} $ but it also seems like a dead end.

In the other way that I tried, I did found an antideriviative of the integrand, but not in the segment

$ [0,\frac{\pi}{2}] $

because if we could divide by $ \sqrt{\sin x} $ then we'd get:

$ \int\frac{1}{1+\sqrt{\cot x}}dx $ then if we substitue $ \sqrt{\cot x}=t $ then

$ \sqrt{\cot x}=t $

so

$ \frac{1}{2t}\cdot\frac{-1}{\sin^{2}x}dx=dt $

and since $ \frac{1}{\sin^{2}x}=1+\cot^{2}x $ we would get

$ dx=\frac{-2t}{1+t^{4}}dt $

Thus

$ \int\frac{1}{1+\sqrt{\cot x}}dx=-\int\frac{2t}{\left(1+t\right)\left(1+t^{4}\right)}dt=-\int\frac{2t}{\left(1+t\right)\left(t^{2}-\sqrt{2}t+1\right)\left(t^{2}+\sqrt{2}t+1\right)}dt=-\int\left(\frac{-1}{t+1}+\frac{1+\sqrt{2}}{2\left(t+\sqrt{2}\right)}+\frac{1-\sqrt{2}}{2\left(t-\sqrt{2}\right)}\right)dt $

and finally:

$ \int\frac{\sqrt{\sin x}}{\sqrt{\sin x}+\sqrt{\cos x}}dx=\ln|\sqrt{\cot x}+1|-\frac{1+\sqrt{2}}{2}\ln|\sqrt{\cot x}+\sqrt{2}|-\frac{1-\sqrt{2}}{2}\ln|\sqrt{\cot x}-\sqrt{2}|+constant $

So this is an antideriviative, but we cannot use Newton leibnitz's formula because of the point $ x=0 $.

In addition, I tried to calculate this integral with an online integral calculator and it failed to show the steps, so I guess this calculation isnt trivial.

Any suggestions would help.

Thanks in advance.

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    $\begingroup$ This is very simple in fact...Try to prove this property $$\int_a^bf(x)dx=\int_a^bf(a+b-x)dx$$ $\endgroup$
    – user35508
    Aug 22, 2020 at 14:09
  • $\begingroup$ @user35508 Is it true without any demands on $ f $ ? how does it helps here ? It would just replace $ sin $ by $ cos $ $\endgroup$
    – FreeZe
    Aug 22, 2020 at 14:13
  • $\begingroup$ Use $y=\frac{\pi}{2}-x$ Substitution $\endgroup$
    – DARK
    Aug 22, 2020 at 14:14
  • $\begingroup$ @Waizman: The integrand should be Reimann integrable on $[a,b]$ which is clearly the case here as the integrand is continuous on $[a,b]$ $\endgroup$
    – Koro
    Aug 22, 2020 at 14:18

3 Answers 3

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Let $I=\intop_{0}^{\frac{\pi}{2}}\frac{\sqrt{\sin x}}{\sqrt{\sin x}+\sqrt{\cos x}}dx\tag{1}$
Substitute $x=\pi/2-t$ so that $dx=-dt$. Hence, $I=-\intop_{\pi/2}^{0}\frac{\sqrt{\cos y }}{\sqrt{\sin y}+\sqrt{\cos y}}dy=\intop_{0}^{\frac{\pi}{2}}\frac{\sqrt{\cos y}}{\sqrt{\sin y}+\sqrt{\cos y}}dy=\intop_{0}^{\frac{\pi}{2}}\frac{\sqrt{\cos x}}{\sqrt{\cos x}+\sqrt{\sin x}}dx\tag{2}$

Add the two to get: $2I= \int_0^{\frac{\pi}{2}} \frac{\sqrt{\sin{x}}+\sqrt{\cos{x}}}{\sqrt{\sin{x}}+\sqrt{\cos{x}}} \; dx \\ = \int_{0}^{\pi/2}dx=\pi/2\implies I=\pi/4$

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  • $\begingroup$ That's brilliant solution, I wouldnt think about it on my own. Thanks $\endgroup$
    – FreeZe
    Aug 22, 2020 at 14:20
  • $\begingroup$ @Waizman, hint was already given to you by -@user35508. $\endgroup$
    – SarGe
    Aug 22, 2020 at 14:39
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As user35508 mentioned in the comments, you can use the substitution of $u=\frac{\pi}{2}-x$ to obtain $$\int_0^{\frac{\pi}{2}} \frac{\sqrt{\cos{x}}}{\sqrt{\sin{x}}+\sqrt{\cos{x}}} \; dx$$ This integral is equivalent to the original integral, so if you add the original integral to the previous integral you get: \begin{align*} 2I&=\int_0^{\frac{\pi}{2}} \frac{\sqrt{\sin{x}}+\sqrt{\cos{x}}}{\sqrt{\sin{x}}+\sqrt{\cos{x}}} \; dx \\ &= \frac{\pi}{4} \end{align*}

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Hint:use kings property ie replace $x$ by $\frac{\pi}{2}-x$ you get an equivalent integral . Now add these integrals what do you observe?

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