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

  • Thread starter Thread starter karush
  • Start date Start date
  • Tags Tags
    Radical Trig
karush
Gold Member
MHB
Messages
3,240
Reaction score
5
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...
 
Physics news on Phys.org
note …

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

also, $\sqrt{\cos^2{t}} = |\cos{t}|$
 
Last edited by a moderator:
skeeter said:
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 -
 
karush said:
$I=\displaystyle\int_{-\pi}^{\pi} |\cos{t}| dt$
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 [math]( -\pi , \pi ][/math]?

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

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

$2 \displaystyle \int_{-\pi/2}^{\pi/2} |\cos{t}| \, dt = 4 \int_0^{\pi/2} \cos{t} \, dt$
 
$$$$
skeeter said:
$\displaystyle \int_{-\pi}^\pi \sqrt{\dfrac{1+\cos{x}}{2}} \, dx$

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

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

$ \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 [math]\left [ -\pi, -\dfrac{ \pi }{2} \right ][/math] and [math]\left [ \dfrac{ \pi }{2}, \pi \right ][/math]. There are 4 intervals which have the same value and are all the same as the integral from 0 to [math]\pi / 2[/math].

-Dan
 
Mahalo
spicy problem 🤔
 
  • #10
karush said:
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...
 

Similar threads

Replies
29
Views
4K
Replies
2
Views
1K
Replies
5
Views
1K
Replies
2
Views
846
Replies
1
Views
2K
Back
Top