Question

integral of sqrt(1-cosx)

Answer 1

http://www.wolframalpha.com/input/?i=integrate+sqrt%281-cosx%29

Answer 2

i could have done it myself, i don't understand how it did it in the steps ...

Answer 3

use the equation
\[\sin ^{2} + \cos^{2} = 1 \to \sqrt{1 - \cos(x)} = \frac{\sin(x)}{\sqrt{1+\cos(x)}} = \frac{\sin(x)}{2 - (1 - cos(x))}\]

Set u = 1-cos(x) so du = sinx dx
giving us \[dx = \frac{du}{sin(x)}\]

Plugging all this in we get \[\int\limits \frac{\sin(x)}{\sqrt{2 - u}} \frac{du}{\sin(x)}\]

Cancel out and do one more u substitution, integrate and then plug back in to get

\[-2\sqrt{1 + cos(x)} + C\]

Answer 4

you could also start out by multiplying by sin/sin to get (faster since you only do one u sub)

\[\int \sqrt{1-cos(x)} * \frac{sin(x)}{sin(x)}dx = \int \sqrt{1-cos(x)} * \frac{sin(x)}{\sqrt{1 - cos(x)}\sqrt{1+cos(x)}}dx\]

and using u = 1 + cos(x)