How to compute this integral involving trig?

Question

$\int_0^{2\pi} \sqrt{2-{2\cos{t}}} \ dt$

This is the integral that I would like to solve. I was told to use the double angle formula for cosine; however, I haven't been able to connect the dots on how to do so.

This integral was simplified from a complex integral.

Thank you!

Answer 1

Your integral equals \begin{align} &=\int_0^{2\pi}\sqrt{2-2\cos t}\,dt\\ &=\sqrt{2}\int_0^{2\pi}\sqrt{1-\cos t}\,dt \\ &=\sqrt{2} \int_0^{2\pi} \sqrt{2}\sin \frac{t}{2} \,dt \\ \tag{1} &=2\int_0^{2\pi} \sin \frac{t}{2} \, dt \\ &=-4\cos\frac{t}{2} \,\,\Bigg|_0^{2\pi}\\ &=8 \end{align} Where I have used the half-angle formula for $\sin$ in (1): $$\sin\frac{\theta}{2}=\sqrt{\frac{1-\cos \theta}{2}}$$

Answer 2

Guide:

To help you continue.

$$\cos t = 1 - 2\sin^2 \left( \frac{t}{2}\right)$$

$$\sqrt{2-2\cos t} =\sqrt{ 2-2\left(1 - 2\sin^2 \left( \frac{t}{2}\right)\right)}=2\sqrt{\sin^2\left(\frac{t}{2} \right)}$$

Answer 3

$$1-\cos t=2 \sin^2 t/2$$ $$\int_0^{2\pi} \sqrt{2-{2\cos{t}}} \ dt=\int_0^{2\pi} 2 \sin(t/2) \ dt=\int_0^{\pi} 4 \sin(t) \ dt=8$$