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

Evaluate

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

I have tried substitution of $$\sin x$$ as well as $$\cos x$$ but it is not giving an answer. Do not understand if there is a formula for this or not.

• Are you sure it's not definite ? Commented Feb 5, 2020 at 18:30
• Please use the Latex parser to type in mathematical formulas. Commented Feb 5, 2020 at 18:43
• Maple does it as an elliptic integral. Commented Feb 5, 2020 at 19:03
• @GEdgar Yet, ultimately the integral can be done with elementary functions. This might be an example of a problem where doing it by hand is actually more efficient than the computer at handling the simplifications. Commented Feb 5, 2020 at 20:48
• I strongly suspect this is a definite integral problem (with bounds of $0$ and $\frac{\pi}2$ probably). While it is instructive and gratifying to see even "difficult" looking indefinite integrals having elementary function expressions, if (and only if) the asker has misrepresented the question by omitting that crucial detail, he should be admonished not to waste others' time and effort in future. Anyway, let's see if there's an update. Commented Feb 6, 2020 at 7:03

\begin{align} I&=\int \frac{\sqrt{\sin x}}{\sqrt{\sin x} + \sqrt{\cos x}} dx\\ &=\int \frac{1}{\sqrt{\cot x}+1}dx =\frac12x -\frac12 \int \frac{\sqrt{\cot x}-1}{\sqrt{\cot x}+1}dx\tag1 \\ \end{align} where, with $$t=\sqrt{\cot x}$$ \begin{align} \int \frac{\sqrt{\cot x}-1}{\sqrt{\cot x}+1}dx &=\int \frac{2t(1-t)}{(1+t)(1 + t^4)}dt\\ &= 2 \int \frac{t^3}{1+t^4}-\frac1{1+t}+\frac{1-t^2}{1+t^4}\ dt\\ &= \frac12\ln(1+t^4)-2\ln(1+t)+2\int \frac{d(t+\frac1t)}{2-(t+\frac1t)^2}dt\\ &= 2\ln \frac{\sqrt[4]{1+t^4}}{1+t}+\sqrt2\coth^{-1}\frac{1+t^2}{\sqrt2t} \end{align}

Substitute into (1) to obtain \begin{align} I= \frac12x+\ln (\sqrt{\sin x}+\sqrt{\cos x}) -\frac1{\sqrt2}\coth^{-1}\frac{\sqrt{\tan x}+\sqrt{\cot x}}{\sqrt{2}} \end{align}

Recall that \begin{align}\sin(x) &= \frac{\tan(x)}{\sec(x)} \\ \cos(x) &= \frac1{\sec(x)} \\ \sec^2(x) &= 1 + \tan^2(x)\end{align} Then, $$\int\frac{\sqrt{\sin(x)}}{\sqrt{\sin(x)} + \sqrt{\cos(x)}}\,\mathrm dx\equiv\int\sec^2(x)\cdot\frac{\sqrt{\tan(x)}}{\sqrt{\tan(x)} \left(1 + \tan^2(x)\right) + 1 + \tan^2(x)}\,\mathrm dx$$ Now, let $$u = \sqrt{\tan(x)}\implies2\cdot\mathrm du = \dfrac{\sec^2(x)}{\sqrt{\tan(x)}}\,\mathrm dx$$. So, \begin{align}\int\sec^2(x)\cdot\frac{\sqrt{\tan(x)}}{\sqrt{\tan(x)} \left(1 + \tan^2(x)\right) + 1 + \tan^2(x)}\,\mathrm dx&\equiv\int2\cdot\frac{u^2}{u\left(1 + u^4\right) + 1 + u^4}\,\mathrm du \\ &= 2\int\frac{u^2}{(1 + u)(1 + u^4)}\,\mathrm du\end{align} From here on, partial fractions and linearity are your friends.

Let $$x=\tan ^{-1}(t)$$ to make $$I=\int \frac{\sqrt{\sin x}}{\sqrt{\sin x} + \sqrt{\cos x}} dx=\int \frac{dt}{t^{3/2}+t^2+\frac{1}{\sqrt{t}}+1}$$ Now, $$t=u^2$$ $$I=\int \frac{2 u^2}{u^5+u^4+u+1}\,du=\int \frac{du}{u+1}-\int \frac{u^3-u^2-u+1}{u^4+1}$$ For the second integral, use the roots of $$u^4+1=0$$ and partial fraction decomposition to face again a series of integrals $$J_k=\int \frac {du}{u+k}=\log(u+k)$$ where $$k$$ is a complex number.

Recombine everything together to get $$I=-\frac{1}{2} \tan ^{-1}\left(\frac{1}{u^2}\right)+\frac{1}{2 \sqrt{2}}\log \left(\frac{u^2-\sqrt{2} u+1}{u^2+\sqrt{2} u+1}\right)+\log \left(\frac{u+1}{\sqrt[4]{u^4+1}}\right)$$

HINT

If $$f(x)=\dfrac{\sqrt{\sin{x}}}{\sqrt{\sin{x}}+\sqrt{\cos{x}}}$$ then $$f(x)+f(\frac{\pi}{2}-x)=1$$