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How to integrate : $$\int\sqrt{\sin x} \cos^{\frac{3}{2}}x dx$$

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Wolfram may help. – Ben Feb 23 '13 at 14:36
Change variable $y=\cos x$ to get $\int(y^2-y^4)^{1/4}dy$, indeed suggesting it is an elliptic integral. – GEdgar Feb 23 '13 at 14:55

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.

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Not sure I understand the game with the fractional powers of cosine... The primitive only exists on an interval where both sine and cosine are nonnegative anyway hence all these precautions seem moot. (+1 for the answer.) – Did Feb 28 '13 at 14:13
@Did The primitive exists in the whole complex plane, assuming we assign the differential constant a value at zeros of $\cos(x)$. The primitive will not be real though. – Sasha Feb 28 '13 at 14:19
To me it seemed obvious that the question was about a primitive on the real line. The first problem with your suggestion is how to define $\sqrt{\sin}$ on the whole complex plane. You need this since the function involves $\sqrt{\sin}$, but already in a neighborhood of zero this is impossible. – Did Feb 28 '13 at 14:26
@Did If the primitive is of interest only for $0\leqslant x \leqslant \frac{\pi}{2}$ then I agree with you. – Sasha Feb 28 '13 at 14:35
Right. (But my stronger claim is that the question makes no sense if not in a real interval $2n\pi+[0,\pi/2]$, $n$ an integer.) – Did Feb 28 '13 at 14:41

$$\begin{aligned} \int \sqrt{\sin x} \cos^{3/2}x\,\mathrm{d}x &=\int\sqrt{\sin x\cos x}\cos x\,\mathrm{d}x\\ &=\frac{1}{\sqrt{2}}\int\sqrt{\sin 2x}\cos x\,\mathrm{d}x\\ &\overset{(1)}{=}\frac{1}{\sqrt{2}}\left(\sin x\sqrt{\sin 2x} - \int\sin x\frac{\cos 2x}{\sqrt{\sin 2x}}\,\mathrm{d}x\right)\\ &=\frac{\sin x\sqrt{\sin 2x}}{\sqrt{2}}-\frac{1}{2}\int\sqrt{\tan x} \cos 2x\,\mathrm{d}x\\ &\overset{(2)}{=}\frac{\sin x\sqrt{\sin 2x}}{\sqrt{2}}-\frac{1}{2}\left(\frac{\sin 2x \sqrt{\tan x}}{2}-\frac{1}{2}\int\sin 2x \frac{\sec^2 x}{2\sqrt{\tan x}}\,\mathrm{d}x\right)\\ &= \frac{\sin x\sqrt{\sin 2x}}{\sqrt{2}}-\frac{\sin 2x \sqrt{\tan x}}{4} + \frac{1}{4}\int \sqrt{\tan x}\,\mathrm{d}x \end{aligned}$$ 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, $$ \int \sqrt{\sin x} \cos^{3/2} x\,\mathrm{d}x = \frac{\sin x\sqrt{\sin 2x}}{\sqrt{2}}-\frac{\sin 2x \sqrt{\tan x}}{4} + \frac{\sqrt{2}}{8}\arctan\left(\sqrt{2\tan x}+1\right)+\frac{\sqrt{2}}{8}\arctan\left(\sqrt{2\tan x}-1\right) - \dfrac{1}{8\sqrt{2}}\ln\!\left(\frac{\tan x+\sqrt{2\tan x}+1}{\tan x-\sqrt{2\tan x}+1}\right) + C $$


$(1)$ – 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: $$\begin{aligned} \int \frac{u^2+1}{u^4+1}\,\mathrm{d}u &= \int\frac{1+1/u^2}{\left(u-1/u\right)^2 + 2}\,\mathrm{d}u\\&=\frac{1}{\sqrt{2}}\arctan\!\left(\frac{u - 1/u}{\sqrt{2}}\right)+ C \end{aligned}$$

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If you're not looking for a closed form, you can use a series expansion and integrate it termwise.

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This integral represented to elementary function. (But this function is so complicated.) See Here. – Hanul Jeon Feb 23 '13 at 14:33
I'm not sure whether a hypergeometric function would qualify as an "elementary" one... – DonAntonio Feb 23 '13 at 14:38
Maple does it in terms of the elliptic integral $\Pi$. Suggesting that it is not elementary. – GEdgar Feb 23 '13 at 14:50

Here is a cleaner form of the solution in terms of the hypergeometric function

$$\int\sqrt{\sin x} \cos^{\frac{3}{2}}x\, dx= \frac{2}{3}\, \sin^{\frac{3}{2}}(x)\, {_2F_1\left(-\frac{1}{4},\frac{3}{4};\,\frac{7}{4};\, \sin^{3}\left( x \right) \right) }+c$$

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There is a typo in your answer. $\sin^3(x)$ should be $\sin^2(x)$ for this answer to be correct at any point. Also, assuming the typo corrected, differentiating it gives the integrand multiplied by a differential constant which is not equal to 1, but it is a piecewise constant function. – Sasha Feb 28 '13 at 14:08

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