# Thread: 2 integrals I dont know how to start

1. ## 2 integrals I dont know how to start

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

2. I think for Q1 can be started by using partial fractions.

3. For the first try partial fractions

$\dfrac{1}{x^6-1} = \dfrac{1}{6}\, \dfrac{1}{ x-1} - \dfrac{1}{6}\, \dfrac{1}{ x+1}-\dfrac{1}{6}\,{
\dfrac {x+2}{{x}^{2}+x+1}}+\dfrac{1}{6}\,{\dfrac {x-2}{{x}^{2}-x+1}}$

and the second, try using the identity

$\cos ax = 1 - 2 \sin^2 \frac{ax}{2}$.

4. im stuck on how u got that partial fraction decomposition?

5. $(x^6-1) =$

$(x^2-1)(x^4+x^2+1) =$

$(x-1)(x+1)(x^4+x^2+1)$

and then

$(x-1)(x+1)(x^2-x+1)(x^2+x+1)$

6. so far with this i have

1/x^6-1 = A/x-1 + B/x+1 + Cx+D/x^2+x+1 + Ex+F/x^2-x+1

I have found A = 1/6, B = -1/6, D = -2, and F = 4/3

I dont know if they are right but I am finding it difficult to work out the other constants.

7. the answer I get for Q2 is:

2*sqrt((cos(ax))+1)/a + C

can any1 verify this?