Integrating :$\int\sqrt{\sin x} \cos^{\frac{3}{2}}x dx$ How to integrate :
$$\int\sqrt{\sin x} \cos^{\frac{3}{2}}x dx$$
 A: Let's make a change of variables $u = \sin^2(x)$. Formally, $\sqrt{\sin(x)} = u^{1/4}$, $\cos^{3/2}(x) = (1-u)^{3/4}$, and $\mathrm{d}x = \frac{\mathrm{d}u}{2 \sqrt{u} \sqrt{1-u}}$.
Thus:
$$
    \int \sqrt{\sin( x)} \cos^{3/2}( x) \, \mathrm{d}x = \frac{1}{2}\int u^{-1/4} (1-u)^{1/4} \mathrm{d} u
$$
In another answer of mine I show how to use differentiation properties of the Gauss's hypergeometric function ${}_2F_1$ to evaluate:
$$
    \int \left(1-u\right)^a u^b \mathrm{d}u = \frac{u^{b+1}}{b+1} {}_2 F_1\left( \left. \begin{array}{cc} -a & b+1 \cr &b+2& \end{array} \right| u \right) +\color\gray{\text{const.}}
$$
Using the above for $b=-1/4$ and $a=1/4$:
$$
  \int u^{-1/4} (1-u)^{1/4} \mathrm{d} u = \frac{4}{3} u^{3/4} \cdot {}_2 F_1\left( \left. \begin{array}{cc} -1/4 & 3/4 \cr &7/4& \end{array} \right| u \right) +\color\gray{\text{const.}}
$$
Recombining we get:
$$
  \int \sqrt{\sin( x)} \cos^{3/2}( x) \, \mathrm{d}x = \frac{2}{3} \sin^{3/2}(x) \cdot {}_2 F_1\left( \left. \begin{array}{cc} -1/4 & 3/4 \cr &7/4& \end{array} \right| \sin^2(x)\right) +\color\gray{\text{const.}} \tag{$\ast$}
$$
Since we use formal operation, like $\sqrt{\sin(x)} = \sqrt{\sqrt{u}} \stackrel{?}{=} u^{1/4}$ we should differentiate $(\ast)$ to check the result. Differentiating we get:
$$
  \frac{\mathrm{d}}{\mathrm{d} x} \left( \frac{2}{3} \sin^{3/2}(x) \cdot {}_2 F_1\left( \left. \begin{array}{cc} -1/4 & 3/4 \cr &7/4& \end{array} \right| \sin^2(x)\right) \right) = \sqrt{\sin(x)} \cos(x) \left(\cos^2(x)\right)^{1/4} 
$$
The above is different from the original integrand by a factor of $\frac{(\cos^2(x))^{1/4}}{\sqrt{\cos(x)}}$ which is a differential constant (whose fourth power simplifies to 1, and which equals 1 where $\cos(x)>0$), and hence we can adjust the $(\ast)$ by simply dividing over it, giving:
$$
   \int \sqrt{\sin( x)} \cos^{3/2}( x) \, \mathrm{d}x = \frac{2}{3} \frac{\sqrt{\cos(x)} \, \sin^{3/2}(x)}{(\cos^2(x))^{1/4}}  \cdot {}_2 F_1\left( \left. \begin{array}{cc} -1/4 & 3/4 \cr &7/4& \end{array} \right| \sin^2(x)\right)+\color\gray{\text{const.}} 
$$
which makes us realize that the constant of integration means a differential constant.
A: If you're not looking for a closed form, you can use a series expansion and integrate it termwise.
A: Here is a cleaner form of the solution in terms of the hypergeometric function
$$\int\sqrt{\sin x} \cos^{\frac{3}{2}}x\, dx= \frac{2}{3}\,  \sin^{\frac{3}{2}}(x)\,  
{_2F_1\left(-\frac{1}{4},\frac{3}{4};\,\frac{7}{4};\, \sin^{3}\left( x \right) \right) }+c$$ 
A: $$\begin{aligned} 
\int \sqrt{\sin x} \cos^{3/2}x\,\mathrm{d}x &=\int\sqrt{\sin x\cos x}\cos x\,\mathrm{d}x\\
&=\frac{1}{\sqrt{2}}\int\sqrt{\sin 2x}\cos x\,\mathrm{d}x\\
&\overset{(1)}{=}\frac{1}{\sqrt{2}}\left(\sin x\sqrt{\sin 2x} - \int\sin x\frac{\cos 2x}{\sqrt{\sin 2x}}\,\mathrm{d}x\right)\\
&=\frac{\sin x\sqrt{\sin 2x}}{\sqrt{2}}-\frac{1}{2}\int\sqrt{\tan x} \cos 2x\,\mathrm{d}x\\
&\overset{(2)}{=}\frac{\sin x\sqrt{\sin 2x}}{\sqrt{2}}-\frac{1}{2}\left(\frac{\sin 2x \sqrt{\tan x}}{2}-\frac{1}{2}\int\sin 2x \frac{\sec^2 x}{2\sqrt{\tan x}}\,\mathrm{d}x\right)\\
&= \frac{\sin x\sqrt{\sin 2x}}{\sqrt{2}}-\frac{\sin 2x \sqrt{\tan x}}{4} + \frac{1}{4}\int \sqrt{\tan x}\,\mathrm{d}x
\end{aligned}$$
To evaluate $\displaystyle{\mathcal{B}=\int\sqrt{\tan x}\,\mathrm{d}x}$, first set $u = \sqrt{\tan x}$ (or $x = \arctan\left(u^2\right)$) and $\mathrm{d}x=\dfrac{2u}{1 + u^4}\,\mathrm{d}u$ to get
$$
\begin{aligned}
\int\frac{2u^2}{u^4 + 1}\,\mathrm{d}u&=\color{red}{\int\frac{u^2 + 1}{u^4 + 1}\,\mathrm{d}u} + \int\frac{u^2 - 1}{u^4 + 1}\,\mathrm{d}u\\
&=\frac{1}{2}\int\frac{\mathrm{d}u}{\left(u + 1/\sqrt{2}\right)^2 + \left(1/\sqrt{2}\right)^2} + \frac{1}{2}\int\frac{\mathrm{d}u}{\left(u - 1/\sqrt{2}\right)^2 + \left(1/\sqrt{2}\right)^2}+\underbrace{\int\frac{1-\dfrac{1}{u^2}}{\left(u + \dfrac{1}{u}\right)^2 - 2}\,\mathrm{d}u}_{\text{Set }w=u+1/u\text{ and }\mathrm{d}w=\left(1-1/u^2\right)\,\mathrm{d}u}\\
&=\frac{\sqrt{2}}{2}\arctan\left(u\sqrt{2}+1\right)+\frac{\sqrt{2}}{2}\arctan\left(u\sqrt{2}-1\right) + \int\frac{\mathrm{d}w}{w^2 - 2}\\
&=\frac{\sqrt{2}}{2}\arctan\left(u\sqrt{2}+1\right)+\frac{\sqrt{2}}{2}\arctan\left(u\sqrt{2}-1\right)-\dfrac{1}{2\sqrt{2}}\ln\left|\dfrac{w+\sqrt{2}}{w-\sqrt{2}}\right| + C_0\\
&\overset{(3)}{=}\frac{\sqrt{2}}{2}\arctan\left(u\sqrt{2}+1\right)+\frac{\sqrt{2}}{2}\arctan\left(u\sqrt{2}-1\right) - \dfrac{1}{2\sqrt{2}}\ln\!\left(\frac{u^2+u\sqrt{2}+1}{u^2-u\sqrt{2}+1}\right)+C_1\\
&=\frac{\sqrt{2}}{2}\arctan\left(\sqrt{2\tan x}+1\right)+\frac{\sqrt{2}}{2}\arctan\left(\sqrt{2\tan x}-1\right) - \dfrac{1}{2\sqrt{2}}\ln\!\left(\frac{\tan x+\sqrt{2\tan x}+1}{\tan x-\sqrt{2\tan x}+1}\right)+C_1
\end{aligned}
$$
Then,
$$ \int \sqrt{\sin x} \cos^{3/2} x\,\mathrm{d}x = \frac{\sin x\sqrt{\sin 2x}}{\sqrt{2}}-\frac{\sin 2x \sqrt{\tan x}}{4} + \frac{\sqrt{2}}{8}\arctan\left(\sqrt{2\tan x}+1\right)+\frac{\sqrt{2}}{8}\arctan\left(\sqrt{2\tan x}-1\right) - \dfrac{1}{8\sqrt{2}}\ln\!\left(\frac{\tan x+\sqrt{2\tan x}+1}{\tan x-\sqrt{2\tan x}+1}\right) + C $$
Notes:
$(1)$ – Integrate by parts with $u=\sqrt{\sin 2x}$ and $\mathrm{d}v = \cos x\,\mathrm{d}x$
$(2)$ – Integrate by parts with $u=\sqrt{\tan x}$ and $\mathrm{d}v = \cos 2x\,\mathrm{d}x $
$(3)$ – I've removed the absolute value bars because the argument is always positive. It is easier to see if you complete the square.
$\color{red}{(*)}$ – You could use the same technique applied to the integral at right, but, in this case, this trick, although clever, leads to a discontinuity in the primitive (at $u=0$). Look:
$$\begin{aligned} \int \frac{u^2+1}{u^4+1}\,\mathrm{d}u &= \int\frac{1+1/u^2}{\left(u-1/u\right)^2 + 2}\,\mathrm{d}u\\&=\frac{1}{\sqrt{2}}\arctan\!\left(\frac{u - 1/u}{\sqrt{2}}\right)+ C \end{aligned}$$
