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

sax
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4 Answers4

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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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Sasha
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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
  • @Did I disagree with your stronger claim. The integrand $f(z)$ can be viewed as a holomorphic function in some region of the complex plane, and one seeks a holomorphic primitive function $F(z)$ such that $F'(z) = f(z)$. This makes perfect sense to me. Am I not following your stronger claim? – Sasha Feb 28 '13 at 15:04
  • Right, I was too hasty, let me rephrase: (1) to me, the question is almost surely meant on (an interval of) the real line, (2) to solve it on (some subset of) the complex plane (as you seem to want to be doing), one must first define precisely said subset, on which holomorphic square roots of sine and cosine must exist. – Did Feb 28 '13 at 15:30
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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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Here is a cleaner form of the solution in terms of the hypergeometric function

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

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

user149844
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