I have these two integrals I need to evaluate but not sure how to start them.
Can any1 show me?
1. integral 1/(x^6-1) dx
2. integral sqrt(1-cos(ax)) dx
For the first try partial fractions
$\displaystyle \dfrac{1}{x^6-1} = \dfrac{1}{6}\, \dfrac{1}{ x-1} - \dfrac{1}{6}\, \dfrac{1}{ x+1}-\dfrac{1}{6}\,{
\dfrac {x+2}{{x}^{2}+x+1}}+\dfrac{1}{6}\,{\dfrac {x-2}{{x}^{2}-x+1}}$
and the second, try using the identity
$\displaystyle \cos ax = 1 - 2 \sin^2 \frac{ax}{2}$.