# difficult integral

• August 26th 2010, 10:21 AM
miranda
difficult integral
hey this integral is really bothering me, the main bit being the k in the middle of it. thanks to anyone who can help :)

0∫π 1/((k + cosx)^2) dx where k>1

cheers!
• August 26th 2010, 12:20 PM
DeMath
Quote:

Originally Posted by miranda
hey this integral is really bothering me, the main bit being the k in the middle of it. thanks to anyone who can help :)

0∫π 1/((k + cosx)^2) dx where k>1

cheers!

$\displaystyle{\int\limits_0^\pi\dfrac{dx}{(k+\cos{ x})^2}.}$

$\dfrac{1}{(k+\cos{x})^2}=\dfrac{1}{\left(k\sin^2\d frac{x}{2}+k\cos^2\dfrac{x}{2}+\cos^2\dfrac{x}{2}-\sin^2\dfrac{x}{2}\right)^2}=$

$=\dfrac{1}{\left((k-1)\sin^2\dfrac{x}{2}+(k+1)\cos^2\dfrac{x}{2}\right )^2}=\dfrac{1}{(k+1)^2}\dfrac{1}{\cos^4\dfrac{x}{2 }\left(\dfrac{k-1}{k+1}\tan^2\dfrac{x}{2}+1\right)^2}=$

$=\dfrac{1}{(k+1)^2}\dfrac{1+\tan^2\dfrac{x}{2}}{\l eft(a\tan^2\dfrac{x}{2}+1\right)^2}\dfrac{1}{\cos^ 2\dfrac{x}{2}},~a=\dfrac{k-1}{k+1}.$

Substitution

$\tan\dfrac{x}{2}=u~\Rightarrow~\dfrac{dx}{\cos^2\d frac{x}{2}}=2\,du$
• August 26th 2010, 12:42 PM
DeMath
So we have

$\displaystyle{\int\limits_0^\pi\dfrac{dx}{(k+\cos{ x})^2}=\dfrac{2}{(k+1)^2}\mathop{\lim}\limits_{b\t o\infty}\int\limits_0^b\dfrac{1+u^2}{(au^2+1)^2}\, du=\ldots=}$

$=\left.{\dfrac{2}{(k+1)^2}\mathop{\lim}\limits_{b\ to\infty}\left(\dfrac{a-1}{2a}\dfrac{u}{au^2+1}+\dfrac{a+1}{2a^{3/2}}\arctan\!\left(\sqrt{a}\,u\right)\right)}\right |_0^b=$

$=\left.{\dfrac{2}{(k+1)^2}\mathop{\lim}\limits_{b\ to\infty}\left(\dfrac{1}{1-k}\dfrac{u}{au^2+1}+\dfrac{k(k+1)^{1/2}}{(k-1)^{3/2}}\arctan\!\left(\sqrt{a}\,u\right)\right)}\right |_0^b=$

$=\dfrac{k\,\pi}{(k^2-1)^{3/2}},~k>0.$