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I have to evaluate:

$$\int_{0}^{\pi/2}\frac{\sqrt{\sin x}}{\sqrt{\sin x}+\sqrt{\cos x}}\, \mathrm{d}x. $$ I can't get the right answer! So please help me out!

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  • $\begingroup$ I can't say why, but the result, according to Mathematica is $\pi/4$. $\endgroup$ – Siminore Jul 6 '12 at 8:18
  • $\begingroup$ It seems the indefinite integral here is an elliptic integral. $\endgroup$ – GEdgar Jul 6 '12 at 20:41
  • $\begingroup$ I have changed the formatting of the title so as to make it take up less vertical space -- this is a policy to ensure that the scarce space on the main page is distributed evenly over the questions. See here for more information. Please take this into consideration for future questions. Thanks in advance. $\endgroup$ – GNUSupporter 8964民主女神 地下教會 Mar 12 '18 at 17:06
  • $\begingroup$ @Kns what substitutions did you try? $\endgroup$ – Narasimham Mar 12 '18 at 17:38
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Let $I$ denote the integral and consider the substitution $u= \frac{\pi }{2} - x.$ Then $I = \displaystyle\int_0^{\frac{\pi }{2}} \frac{\sqrt{\cos u}}{\sqrt{\cos u } + \sqrt{\sin u }} du$ and $2I = \displaystyle\int_0^{\frac{\pi }{2}} \frac{\sqrt{\cos u} + \sqrt{\sin u }}{\sqrt{\cos u } + \sqrt{\sin u }} du = \frac{\pi }{2}.$ Hence $I = \frac{\pi }{4}.$

In general, $ \displaystyle\int_0^a f(x) dx = \displaystyle\int_0^a f(a-x) $ $dx$ whenever $f$ is integrable, and $\displaystyle\int_0^{\frac{\pi }{2}} \frac{\cos^a x}{\cos^a x + \sin^a x } dx = \displaystyle\int_0^{\frac{\pi }{2}} \frac{\sin^a x}{\cos^a x + \sin^a x } dx = \frac{\pi }{4}$ for $a>0$ (same trick.)

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  • 1
    $\begingroup$ Is the $f$-continuous condition actually necessary? It seems like the equation you state should be true for any integrable function... $\endgroup$ – Ben Millwood Jul 6 '12 at 10:16
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Note that $\sin(\pi/2-x)=\cos x$ and $\cos(\pi/2-x)=\sin x$. The answer will exploit the symmetry.

Break up the original integral into two parts, (i) from $0$ to $\pi/4$ and (ii) from $\pi/4$ to $\pi/2$. So our first integral is $$\int_{x=0}^{\pi/4} \frac{\sqrt{\sin x}}{\sqrt{\sin x}+\sqrt{\cos x}}\,dx.\tag{$1$} $$

For the second integral, make the change of variable $u=\pi/2-x$. Using the fact that $\sin x=\sin(\pi/2-u)=\cos u$ and $\cos x=\cos(\pi/2-u)=\sin u$, and the fact that $dx=-du$, we get after not much work $$\int_{u=\pi/4}^{0} -\frac{\sqrt{\cos u}}{\sqrt{\cos u}+\sqrt{\sin u}}\,du$$ Change the dummy variable of integration variable to the name $x$. Also, do the integration in the "right" order, $0$ to $\pi/4$. That changes the sign, so our second integral is equal to $$\int_{x=0}^{\pi/4} \frac{\sqrt{\cos x}}{\sqrt{\cos x}+\sqrt{\sin x}}\,dx.\tag{$2$}$$ Our original integral is the sum of the integrals $(1)$ and $(2)$. Add, and note the beautiful cancellation $\frac{\sqrt{\sin x}}{\sqrt{\sin x}+\sqrt{\cos x}}+ \frac{\sqrt{\cos x}}{\sqrt{\cos x}+\sqrt{\sin x}}=1$. Thus our original integral is equal to $$\int_0^{\pi/4}1\,dx.$$ This is trivial to compute: the answer is $\pi/4$.

Remark: Let $f(x)$ and $g(x)$ be any reasonably nice functions such that $g(x)=f(a-x)$. Exactly the same argument shows that $$\int_0^a\frac{f(x)}{f(x)+g(x)}\,dx=\frac{a}{2}.$$

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