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?
Take $u=1+2\cos{x}$
$\frac{-1}{2}du=\sin{x}dx$
So the integral becomes $$\frac{-1}{2}\int \frac{1}{\sqrt{u}}du=-\sqrt{u}+c$$