Computing Int. sqrt(1-cost) | Just an Integral

[quasar987]

How could I compute

$$\int \sqrt{1-cost} \ \ dt$$

?

 

[stunner5000pt]

just throwing this out there... could it have anything to do with a hanfle angle formula??

because i do rmember that
$$\sin t/2 = \pm \sqrt{\frac{1 - \cos (t)}{2}}$$

that plus or minus does make things a bit... iffy though

 

[acm]

Sqrt ( 1- cost)
Sqrt ( 1 - cos^2t) / Sqrt (1 + Cost)
Sqrt ( Sin^2t) / Sqrt (1 + Cost)
Sint / Sqrt (1 + Cost)
Integral becomes: I = Int ( sint/ sqrt( 1+ Cost) ) dt
Let u = 1 + Cost
du/dt= -Sint
du = -Sintdt
Integral becomes: I = -Int ( -du/Sqrt (u) )
I = -2 Sqrt(u) + C
I = -2 Sqrt (1 + Cost) + C

 

[quasar987]

Excellent! Thanks you very much acm!