Question

integral of square root of (1-cos x)dx

Answer 1

\[\int\limits_{?}^{?} \sqrt{1-\cos x} dx\]
is this the question?

Answer 2

Use a substitution \(u=1-\cos{x}\). I think it will work.

Answer 3

yes, and no a u sub won't work du = sinx dx, which we don't have

Answer 4

yes to janie no to mr. math

Answer 5

it is a crazy integral, and I know there is a way to solve it, I just don't know how to get at it.

Answer 6

try wolframalfpha

Answer 7

man, I just couldn't get an answer with the steps shown when I googled wolfram, thanks a bunch janie

Answer 8

oops, nevermind. I forgot the sqrt. you should just ask your teacher - he/she would do a much better job of explaining this problem than me.

Answer 9

How about a trig half angle substitution....if you need more hints letm me know

Answer 10

this one is tricky - i just feel that u = 1-cosx should work - i'll try again

Answer 11

use the substitution for sin(x/2)

Answer 12

it will work out nice.

Answer 13

I'm sure the substitution will work. Here is a hint:
\[1-\cos{x}=u \implies \cos{x}=1-u \implies \sin{x}=\sqrt{2u-u^2}.\]

Answer 14

You can draw a triangle to get that.

Answer 15

So
\[du=\sin{x}dx \implies dx=\frac{du}{\sqrt{2u-u^2}}.\]
If we call our original integral I, then
\[I=\int \frac{\sqrt{u}}{\sqrt{u}\sqrt{2-u}}du=\int \frac{du}{\sqrt{2-u}}.\]

Answer 16

Which is a straight forward integral.

Answer 17

right math - thats the way to go

Answer 18

now use substitution t = 2-u

Answer 19

integral - dt/sqrt t

Answer 20

sorry thats
dt/sqrt t
integral of this is 2 sqrt t
= 2 sqrt (2-u)
= 2 sqrt (cosx + 1) + c

Answer 21

\[\cos{x}=\cos^2{x/2}-\sin^2{x/2}=1-\sin^2{x/2}-\sin^2{x/2}=1-2\sin^2{x/2}\]\[\Rightarrow 1-\cos{x}=2\sin^2{x/2}\]

Answer 22

thanks a ton guys, I had to run out but Mr. Math was totally right on.

Answer 23

pythagorean correlation I didn't take into account for the sin(x)dx, wonderful stuff.