synchcola | Jan. 20th, 2007

You're viewing

synchcola's journal
Create a Dreamwidth Account Learn More

Reload page in style: light

Jan. 20th, 2007

an integral

$\begin{align*} \int_0^{\pi/2} {k' \sin x \over \sqrt{1 - k'^2 \sin^2 x}} \,dx &= \int_0^1 {k' \over \sqrt{k^2 + k'^2 x^2}} \, dx \\&= \mathop{\text{arcsinh}} (k'/k)x \big]_{x=0}^{x=1} \\&= \ln \bigg( {k' \over k} + {1\over k}\bigg) \end{align*}$

(then i went shopping)