# Challenging Integral

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• Jun 21st 2009, 11:55 PM
simplependulum
Challenging Integral
solve $\displaystyle \int \frac{dx}{ sin^5{x} + cos^5{x} }$

and

$\displaystyle \int \frac{dx}{ sin^7{x} + cos^7{x} }$
• Jun 22nd 2009, 06:52 AM
NonCommAlg
Quote:

Originally Posted by simplependulum

solve $\displaystyle \int \frac{dx}{ sin^5{x} + cos^5{x} }$ and $\displaystyle \int \frac{dx}{ sin^7{x} + cos^7{x} }$

i'll solve the first one only and i'll leave the second integral to you or whoever likes trouble! first we need the following magical identity: (Wink)

$\displaystyle \frac{1}{\sin^5 x + \cos^5 x} = \frac{4}{5} \left[\frac{1}{\sin x + \cos x} + \frac{(\sin x + \cos x)((\sin x - \cos x)^2 - 2)}{(\sin x - \cos x)^4 - 4(\sin x - \cos x)^2 - 1} \right].$

now finding $\displaystyle \int \frac{dx}{\sin x + \cos x}$ is straightforward. to find $\displaystyle I= \int \frac{(\sin x + \cos x)((\sin x - \cos x)^2 - 2)}{(\sin x - \cos x)^4 - 4(\sin x - \cos x)^2 - 1} \ dx$ substitute: $\displaystyle \sin x - \cos x = t.$ then $\displaystyle (\sin x + \cos x) \ dx = dt$ and thus

$\displaystyle I=\int \frac{t^2-2}{t^4 - 4t^2 - 1} \ dt.$ finally let $\displaystyle a=2+\sqrt{5}$ and $\displaystyle b=2-\sqrt{5}$ to get: $\displaystyle I=\frac{1}{2} \left[\int \frac{dt}{t^2-a} + \int \frac{dt}{t^2 - b} \right].$ note that $\displaystyle b < 0.$ now everything is nice and easy and you're good to go!

Remark: i think this is true that for any $\displaystyle n \in \mathbb{N}$ we have $\displaystyle \frac{1}{\sin^{2n+1}x + \cos^{2n+1} x} = \frac{2n}{2n+1}\left[\frac{1}{\sin x + \cos x} + (\sin x + \cos x)R((\sin x - \cos x)^2) \right],$ where $\displaystyle R(z)$ is some rational function of $\displaystyle z$.

actually i think $\displaystyle R(z)=\frac{p(z)}{q(z)},$ where both $\displaystyle p(z)$ and $\displaystyle q(z)$ are monic polynomials with integer coefficients! for example, in your problem: $\displaystyle p(z)=z-2, \ q(z)=z^2-4z-1.$ i also think that we

always have $\displaystyle \deg q(z) = 1+\deg p(z).$ these claims are either wrong or easy to prove! (Rofl) ... let's just leave it as a challenge for whoever has time and patience to try it.

by the way, the identity $\displaystyle \frac{1}{\sin^3x + \cos^3x}=\frac{2}{3} \left[\frac{1}{\sin x + \cos x} + \frac{\sin x + \cos x}{(\sin x - \cos x)^2 + 1} \right]$ shows that my claims are not based on nothing! (Evilgrin)
• Jun 22nd 2009, 09:29 PM
simplependulum
My solution is to substitute $\displaystyle t = \sin{x}+\cos{x}$
so,
$\displaystyle dt = (\cos{x} - \sin{x} )dx$

$\displaystyle \sin{x} \cos{x} = \frac{t^2 - 1}{2}$

$\displaystyle dx = \frac{dx}{ \sqrt{ 2 - t^2 } }$

Although it looks very troublesome , it become quite well after separating it into fractions .

However , i think your method is better than what i am using . (Clapping)