How to Evaluate the Integral of an Even Trig Function in a Radical?

[karush]

Evaluate the integral
$I_4=\displaystyle\int_{-\pi}^{\pi}\sqrt{\frac{1+\cos{x}}{2}} \, dx $

ok offhand i think what is in the radical is trig identity
but might be better way...

 

[skeeter]

note …

$\cos^2{t} = \dfrac{1+\cos(2t)}{2}$

also, $\sqrt{\cos^2{t}} = |\cos{t}|$

 

[karush]

note …

$\cos^2{t} = \dfrac{1+\cos(2t)}{2}$

also, $\sqrt{\cos^2{t}} = |\cos{t}|$

$I=\displaystyle\int_{-\pi}^{\pi} |\cos{t}| dt$
Ok I would assume the abs would mean the splitting of the int into 2 int of + and -

 

[topsquark]

$I=\displaystyle\int_{-\pi}^{\pi} |\cos{t}| dt$
Ok I would assume the abs would mean the splitting of the int into 2 int of + and -

Right. So on what intervals is cos(t) positive and negative on \(\displaystyle ( -\pi , \pi ]\)?

-Dan

 

[karush]

Right. So on what intervals is cos(t) positive and negative on \(\displaystyle ( -\pi , \pi ]\)?

-Dan

https://www.desmos.com/calculator/w4iav0jfj1trom are the desmos $[-\pi, 0]$ is - and $[0,\pi]$ is +

 

[skeeter]

$\displaystyle \int_{-\pi}^\pi \sqrt{\dfrac{1+\cos{x}}{2}} \, dx$

$x = 2t \implies dx = 2 \, dt$

$2 \displaystyle \int_{-\pi/2}^{\pi/2} |\cos{t}| \, dt = 4 \int_0^{\pi/2} \cos{t} \, dt$

 

[karush]

$\displaystyle \int_{-\pi}^\pi \sqrt{\dfrac{1+\cos{x}}{2}} \, dx$

$x = 2t \implies dx = 2 \, dt$

$2 \displaystyle \int_{-\pi/2}^{\pi/2} |\cos{t}| \, dt = 4 \int_0^{\pi/2} \cos{t} \, dt$

$ \displaystyle4 \int_0^{\pi/2} \cos{t} \, dt = 4\Biggr[ \sin{t} \Biggr]_0^{\pi/2} =4[1-0]=4$

ok I didn't understand how this last step made the abs disappear?

W|A

 

[topsquark]

$ \displaystyle4 \int_0^{\pi/2} \cos{t} \, dt = 4\Biggr[ \sin{t} \Biggr]_0^{\pi/2} =4[1-0]=4$

ok I didn't understand how this last step made the abs disappear?

W|A

cos(t) is negative on \(\displaystyle \left [ -\pi, -\dfrac{ \pi }{2} \right ]\) and \(\displaystyle \left [ \dfrac{ \pi }{2}, \pi \right ]\). There are 4 intervals which have the same value and are all the same as the integral from 0 to \(\displaystyle \pi / 2\).

-Dan

 

[karush]

Mahalo
spicy problem

 

[Prove It]

Evaluate the integral
$I_4=\displaystyle\int_{-\pi}^{\pi}\sqrt{\frac{1+\cos{x}}{2}} \, dx $

ok offhand i think what is in the radical is trig identity
but might be better way...

It might be worth noting that the integrand is an even function...