1. ## 2 integrals

I have some problems proving that:

$\displaystyle \int_0^{2\pi}$ d$\displaystyle \theta$ / (a+b cos$\displaystyle \theta$) = $\displaystyle 2\pi$ / $\displaystyle \sqrt{a^2-b^2}$, if a>|b| and a,b in $\displaystyle \mathbb{R}$.

$\displaystyle \int_0^1$$\displaystyle \log(\displaystyle \sin$$\displaystyle \pi$x)dx=-$\displaystyle \log $$\displaystyle 2 Thanks. 2. ## Re: 2 integrals Originally Posted by Veve I have some problems proving that: \displaystyle \int_0^{2\pi} d\displaystyle \theta / (a+b cos\displaystyle \theta) = \displaystyle 2\pi / \displaystyle \sqrt{a^2-b^2}, if a>|b| and a,b in \displaystyle \mathbb{R}. \displaystyle \int_0^1$$\displaystyle \log$($\displaystyle \sin$$\displaystyle \pix)dx=-\displaystyle \log$$\displaystyle 2$

Thanks.
For the first you need to use the Weierstrauss Substitution.

3. ## Re: 2 integrals

Originally Posted by Veve
I have some problems proving that:

$\displaystyle \int_0^{2\pi}$ d$\displaystyle \theta$ / (a+b cos$\displaystyle \theta$) = $\displaystyle 2\pi$ / $\displaystyle \sqrt{a^2-b^2}$, if a>|b| and a,b in $\displaystyle \mathbb{R}$.

$\displaystyle \int_0^1$$\displaystyle \log(\displaystyle \sin$$\displaystyle \pi$x)dx=-$\displaystyle \log$$\displaystyle 2$

Thanks.
For the first one, you can do the following $\displaystyle \displaystyle I=\int_0^{2\pi}\frac{dx}{a+b\cos(x)}=-2i\int_0^{2\pi} \frac{ie^{ix}}{2ae^{ix}+b+be^{2ix}}\;=-2i\oint_{|z|=1}\frac{1}{2az+b+bz^2}$. Now, if $\displaystyle \displaystyle f(z)=\frac{-2i}{2az+b+bz^2}$ you can check that $\displaystyle f$ has a poles at precisely one place on or in the interior of $\displaystyle |z|=1$, namely one at the interior point $\displaystyle \displaystyle z_0=\frac{\sqrt{a^2-b^2}-a}{b}$. We see then by Cauchy's Integral Formula then that $\displaystyle \displaystyle I=2\pi i \,\text{Res}(f,z_0)=\frac{2\pi}{\sqrt{a^2-b^2}}$.

If you look down the solution to this problem you will see I solved precisely the second integral.