Evaluation of $ \int\frac{\sqrt{\sin x}}{\sqrt{\sin x}+\sqrt{\cos x}}dx$

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  • Integral of $\int_0^{\pi/2} \frac {\sqrt{\sin x}}{\sqrt {\sin x} + \sqrt {\cos x}}$dx [duplicate]

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Solutions Collecting From Web of "Evaluation of $ \int\frac{\sqrt{\sin x}}{\sqrt{\sin x}+\sqrt{\cos x}}dx$"

By collecting all the suggestions, you should be able to prove that:
$$\frac{2t}{(t+1)(t^4+1)}=\frac{t}{1+t^4}+\frac{t^3}{1+t^4}+\frac{1-t^2}{1+t^4}-\frac{1}{1+t},$$
and since $(1+t^4)=(1+\sqrt{2}t+t^2)(1-\sqrt{2}t+t^2)$, it follows that:
$$\int\frac{2t\,dt}{(t+1)(t^4+1)}=\frac{1}{2}\arctan t^2+\frac{1}{4}\log(1+t^4)+\frac{1}{2\sqrt{2}}\log\frac{1+\sqrt{2}t+t^2}{1-\sqrt{2}t+t^2}-\log(1+t).$$
$\\$

Addendum

Suggestion for evaluating $\displaystyle{\int \frac{1 – t^2}{t^4 + 1}\,\mathrm{d}t}$:
Write
$$
\begin{aligned}
\int- \frac{t^2 – 1}{t^4 + 1}\,\mathrm{d}t &= -\int \frac{1 – \dfrac{1}{t^2}}{t^2 + \dfrac{1}{t^2}}\,\mathrm{d}t \\
&=-\int \frac{1 – \dfrac{1}{t^2}}{\left(t + \dfrac{1}{t}\right)^2 – 2}\,\mathrm{d}t
\end{aligned}
$$

Now, set $\displaystyle{u = t + \frac{1}{t}}$ and $\mathrm{d}u = \left(1 – \dfrac{1}{t^2}\right)\,\mathrm{d}t$:
$$
\begin{aligned}
-\int \frac{\mathrm{d}u}{u^2 – 2} &= \dfrac{1}{2\sqrt{2}}\ln\left|\frac{u + \sqrt{2}}{u – \sqrt{2}} \right| + C \\
&=\frac{1}{2\sqrt{2}}\ln\left|\frac{t^2 + t\sqrt{2} +1}{t^2 – t\sqrt{2}+1} \right| + C
\end{aligned}
$$
Note that $t^2 + t\sqrt{2}+1 = \left(t + \dfrac{1}{\sqrt{2}}\right)^{\!2} + \dfrac{1}{2} >0$ and $t^2 – t\sqrt{2}+1 = \left(t – \dfrac{1}{\sqrt{2}}\right)^{\!2} + \dfrac{1}{2} >0$, then we can get rid of the absolute value bars:
$$\int \frac{1-t^2}{t^4 + 1}\,\mathrm{d}t = \frac{1}{2\sqrt{2}}\ln\!\left(\dfrac{t^2 + t\sqrt{2}+1}{t^2 – t\sqrt{2}+1}\right) + C $$

This is a very famous technique.