<s>[INST] Verifying the integral solution: $I= \int \frac{\sin x}{\sqrt{1+2\cos x}}dx$ 
Evaluate:
  $$
I= \int \frac{\sin x}{\sqrt{1+2\cos x}}dx
$$

I got an answer which does not match the keys section of the book I'm solving and want to verify which one is correct.
Start by recalling:
$$
\cos 2x = \cos^2 x-\sin^2 x
$$
Therefore:
$$
\sqrt{1+2\cos x} =\sqrt{1+\cos^2 x - \sin^2 x} = \sqrt{2\cos^2 x} = \sqrt2|\cos x|
$$
Thus:
$$
I = \int \frac{\sin x}{\sqrt2 |\cos x|}dx
$$
Substitute $t=\cos x$, then $dt = -\sin x dx$, $dx = - {dt\over \sin x}$, hence:
$$
I = -{1\over \sqrt 2}\int \frac{dt}{|t|} = \boxed{-{\text{sgn}(\cos x)\over \sqrt 2}\ln(|\cos x|)+C}
$$
While the answer section suggests:
$$
I = -\sqrt{1+2\cos x} + C
$$
Which one is correct?
 [/INST] Take $u=1+2\cos{x}$
$\frac{-1}{2}du=\sin{x}dx$
So the integral becomes $$\frac{-1}{2}\int \frac{1}{\sqrt{u}}du=-\sqrt{u}+c$$
</s>