Integrate $\frac{\sin^3 \frac{x}{2}}{\cos \frac{x}{2} \sqrt{\cos x+\cos^2 x+\cos^3 x}}$ Evaluate $$\int \frac{\sin^3 \frac{x}{2}}{\cos \frac{x}{2} \sqrt{\cos x+\cos^2 x+\cos^3 x}}dx$$ I saw terms like $1+t+t^2$ in the denominator , so I thought of $t^3-1$ and then converting back into half angle but it doesn't help me. I am unable to simplify it further. Please tell me a better way to start. Thanks.
 A: Let $$\displaystyle I = \int \frac{\sin ^3(x/2)}{\cos(x/2)\sqrt{\cos^3 x+\cos^2x+\cos x}}dx = \frac{1}{2}\int\frac{2\sin^2 \frac{x}{2}\cdot 2\sin \frac{x}{2}\cdot \cos \frac{x}{2}}{2\cos^2 \frac{x}{2}\sqrt{\cos^3 x+\cos^2 x+\cos x}}dx$$
So we get $$\displaystyle I = \frac{1}{2}\int\frac{(1-\cos x)\cdot \sin x}{(1+\cos x)\sqrt{\cos^3 x+\cos^2 x+\cos x}}dx$$
Now Put $\cos x = t\;,$ Then $\sin x dx = -dt$
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 $$\displaystyle \displaystyle I = \tan^{-1}\sqrt{\left(\cos x+\sec x+1\right)}+\mathcal{C}$$
A: I might be wrong but if you combine these two identities :
$$ \sin^2(x) = 1 - \cos^2(x) $$
$$ \cos(x) = \cos(\frac{x}{2} + \frac{x}{2}) = \cos^2(\frac{x}{2}) - \sin^2(\frac{x}{2}) = 2\cos^2(\frac{x}{2}) - 1 $$
You can expand the expressions until only the sine function in the numerator remains and everything else is powers of cosine.
$$ \sin^3(\frac{x}{2}) = \sin(\frac{x}{2})(1 - \cos^2(\frac{x}{2}))$$
Then if you make the substitution :
$$ t = \cos(\frac{x}{2}) $$
 $$ dt = -\frac{1}{2}\sin(\frac{x}{2})dx $$
It should remove all the trigonometric functions.
I don't know if this will work to solve the integral, but I hope it helps!
