1. ## 2 integrals

I have some problems proving that:

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

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

Thanks.

2. ## Re: 2 integrals

Originally Posted by Veve
I have some problems proving that:

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

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

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:

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

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

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

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