So here is the problem I'm working with
$$\int\sqrt{1-\cos2x}~dx$$
I'm assuming that I'll need to use the trig identity
$2\sin^2x=1-\cos2x$ . But where do I go from there?
$$\int\sqrt{2\sin^2x}~dx$$
Do I split the $\sqrt{2}$ and the $\sqrt{\sin^2x}$ so that I have $$\int\sqrt{2}\sin x~dx$$ and do integration by parts?
Best Answer
Yes, you can simplify as you did at the end, but no need for integration by parts! Recall, $\sqrt 2$ is merely a constant!
$$\int \sqrt 2 \sin x \,dx = \sqrt 2 \int \sin x \,dx = -\sqrt 2 \cos x + C$$
You did the "hardest part" by recognizing the trigonometric identity here. The rest is simply knowing that $\int \sin x = -\cos x + C$