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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
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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

3 Answers 3

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

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. –  tetori 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
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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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2  
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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