Help Integrating Trig

Question

I was doing an FP3 question and got stumped on an integral, had a look at the mark scheme and they made this jump: $$\int_{0}^{2\pi}\sqrt{2-2\cos(x)} dx=2\int_{0}^{2\pi}\sin(\frac{x}{2})dx$$ Could someone please explain this to me because I can't for the life of me see where its come from Thanks :)

Answer 1

You can recall $$ \frac{1-\cos x}{2}=\sin^2\frac{x}{2} $$ If $0\le x\le2\pi$, then $0\le x/2\le\pi$, so $\sin(x/2)\ge0$ and therefore $$ \sqrt{2-2\cos x}=2\sin\frac{x}{2} $$ So your integral becomes $$ 2\int_0^{2\pi}\sin\frac{x}{2}\,dx=2\Bigl[-2\cos\frac{x}{2}\Bigr]_0^{2\pi}=8 $$ The suggested $\pi/2$ is definitely wrong.

Answer 2

HINT: we get $$\cos\left(\frac{x}{2}+\frac{x}{2}\right)=\cos^2\left(\frac{x}{2}\right)-\sin^2\left(\frac{x}{2}\right)$$

Answer 3

When you see something like this, you should check a trigonometric identity table, keeping in mind that the exact form may be slightly different. (i.e., a 1/2 on one side may have been moved to the other as a 2, or a square root on one side may have been moved to the other as a square)

Regardless, this is likely what you are looking for: $2sin^2(x/2) = 1 - cos(x)$. However, with some manipulation, we find $2sin(x/2) = \sqrt{2 - 2cos(x)}$, which is more directly applicable.

The bounds of integration present a different problem. I don't see any way you can get those to make sense. I question your source on that point.