# 7.2.6 trig int with radical

• MHB
• karush

#### karush

Gold Member
MHB
Evaluate
$$\int_0^{2\pi}\sqrt{\dfrac{1-\cos{x}}{2}}\,dx$$
ok my baby step is
$$\int _0^{2\pi }\frac{\sqrt{1-\cos \left(x\right)}}{\sqrt{2}}dx$$
then ?

W|A said the answer was 4

Evaluate
$$\int_0^{2\pi}\sqrt{\dfrac{1-\cos{x}}{2}}\,dx$$
ok my baby step is
$$\int _0^{2\pi }\frac{\sqrt{1-\cos \left(x\right)}}{\sqrt{2}}dx$$
then ?

W|A said the answer was 4
$\cos x = 1 - 2\sin^2\frac x2$

$$I=\int _0^{2\pi }\frac{\sqrt{1-\cos \left(x\right)}}{\sqrt{2}}dx= \int _0^{2\pi }\dfrac{\sqrt{1-\left(1 - 2\sin^2\dfrac x2\right)}}{\sqrt{2}}dx= \int _0^{2\pi }\dfrac{\sqrt{2\sin^2{\dfrac x2}}}{\sqrt{2}}=\int _0^{2\pi }\sin{\frac x2}\,dx$$
then
$$I=-2\cos{\dfrac x2}\biggr| _0^{2\pi }=$$

ok I don't see this approaching 4

$$I=\int _0^{2\pi }\frac{\sqrt{1-\cos \left(x\right)}}{\sqrt{2}}dx= \int _0^{2\pi }\dfrac{\sqrt{1-\left(1 - 2\sin^2\dfrac x2\right)}}{\sqrt{2}}dx= \int _0^{2\pi }\dfrac{\sqrt{2\sin^2{\dfrac x2}}}{\sqrt{2}}=\int _0^{2\pi }\sin{\frac x2}\,dx$$
then
$$I=-2\cos{\dfrac x2}\biggr| _0^{2\pi }=$$

ok I don't see this approaching 4
$$I=-2\cos{\dfrac x2}\biggr| _0^{2\pi }= -2(\cos\pi - \cos0) = -2(-1 - 1) = 4$$