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I came about the following practice problem in a book of integration:-

$Q.$ Evaluate $$I=\int \frac{\sin ^3(\theta/2)}{\cos(\theta/2)\sqrt{\cos^3\theta+\cos^2\theta+\cos \theta}}d\theta$$ To do this, first I substituted $\cos(\theta/2)=u \implies \frac {-1}{2}\sin (\frac{\theta}2)\ d\theta=du \implies \sin^3 (\frac{\theta}2)d\theta=-2(1-u^2)\ du$. This gives $$\begin{align}I&=\int \frac{2(u^2-1)\ du}{u\sqrt{(2u^2-1)^3+(2u^2-1)^2+(2u^2-1)}}\\&=\frac 12\int \frac {(u^2-1)(4u\ du)}{u^2\sqrt{(2u^2-1)^3+(2u^2-1)^2+(2u^2-1)}}\end{align}$$ Now substitute $z=2u^2-1 \implies dz=4u\ du$. We have $u^2=\frac {z+1}2 \implies u^2-1=\frac {z-1}2$. Hence $$\begin{align}I&=\int \frac {{{z-1}\over2}\ dz}{(\frac {z+1}2)\sqrt{z^3+z^2+z}}\\&=\int \frac{(z-1)\ dz}{(z+1)\sqrt{z^3+z^2+z}}\\&=\int \left[\frac{1}{\sqrt{z^3+z^2+z}}-\frac2{(z+1)\sqrt{z^3+z^2+z}}\right]\ dz\end{align}$$ I don't know how to proceed further. Some hints would be appreciated. Is there any easier way to integrate the given expression? I also tried substituting $\tan (\theta/2)=u$, but it becomes even more messier than what I have shown here.

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    $\begingroup$ Just out of curiosity what's the name of the book? $\endgroup$ – MathematicalPhysicist Sep 5 '15 at 13:45
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    $\begingroup$ I have the same request : what is the book which hosts such a monster ? I am almost sure that elliptic functions are in the solution. $\endgroup$ – Claude Leibovici Sep 5 '15 at 14:07
  • $\begingroup$ It is not a standard book that you would find in stores, it is an institute practice book.. $\endgroup$ – Abhishek Bakshi Sep 5 '15 at 14:08
  • $\begingroup$ Mathematica 10.2 gives an answer in terms of elliptic integral. $\endgroup$ – user64494 Sep 5 '15 at 14:24
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Let $$\displaystyle I = \int \frac{\sin ^3(\theta/2)}{\cos(\theta/2)\sqrt{\cos^3\theta+\cos^2\theta+\cos \theta}}d\theta = \frac{1}{2}\int\frac{2\sin^2 \frac{\theta}{2}\cdot 2\sin \frac{\theta}{2}\cdot \cos \frac{\theta}{2}}{2\cos^2 \frac{\theta}{2}\sqrt{\cos^3 \theta+\cos^2 \theta+\cos \theta}}d\theta$$

So we get $$\displaystyle I = \frac{1}{2}\int\frac{(1-\cos \theta)\cdot \sin \theta}{(1+\cos \theta)\sqrt{\cos^3 \theta+\cos^2 \theta+\cos \theta}}d\theta$$

Now Put $\cos \theta = t\;,$ Then $\sin \theta d\theta = -dt$

So Integral $$\displaystyle I = -\frac{1}{2}\int\frac{(1-t)}{(1+t)\sqrt{t^3+t^2+t}}dt = -\frac{1}{2}\int\frac{(1-t^2)}{(1+t)^2\sqrt{t^3+t^2+t}}dt$$

So we get $$\displaystyle I = \frac{1}{2}\int\frac{\left(1-\frac{1}{t^2}\right)}{\left(t+\frac{1}{t}+2\right)\sqrt{t+\frac{1}{t}+1}}dt$$

Now Let $\displaystyle \left(t+\frac{1}{t}+1\right) = u^2\;,$ Then $\left(1-\frac{1}{t^2}\right)dt = 2udu$

So Integral $$\displaystyle I = \frac{1}{2}\int\frac{2u}{u^2+1}\cdot \frac{1}{u}du = \tan^{-1}(u)+\mathcal{C}$$

So we get $$\displaystyle I = \tan^{-1}\sqrt{\left(t+\frac{1}{t}+1\right)}+\mathcal{C}$$

So we get $$\displaystyle \displaystyle I = \tan^{-1}\sqrt{\left(\cos \theta+\sec \theta+1\right)}+\mathcal{C}$$

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  • $\begingroup$ This is pure beauty ! Congratulations and thanks ! $\endgroup$ – Claude Leibovici Sep 5 '15 at 14:29
  • $\begingroup$ actually i have write $-(1-t^2) = (t^2-1)$ $\endgroup$ – juantheron Sep 5 '15 at 14:41
  • $\begingroup$ I see, you found my z-integral much faster. Very nice solution and thanks.. $\endgroup$ – Abhishek Bakshi Sep 5 '15 at 14:53
  • $\begingroup$ Nice solution.BTW @AbhishekBakshi most of such integrals are found by finding the differential coefficient in expression :-) which is wonderfully shown above! $\endgroup$ – user220382 Sep 5 '15 at 15:34
  • $\begingroup$ I see, thanx for the advice.. $\endgroup$ – Abhishek Bakshi Sep 11 '15 at 14:08

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