A closed form for the antiderivative of $\frac 1{\sin^5 x +\cos^5 x} $

Question

Does there exist a closed form expression for $$\int \dfrac {dx}{\sin^5 x +\cos^5 x}? $$

Answer 1

Yes there is. Any rational expression in sine and cosine has an elementary antiderivative. Use the substitution $t=\tan(x/2)$, follow your nose, and you should end up with a rational integrand of $t$. Then partial fraction decomposition will take you home. This general method will work for any rational expression of sines and cosines. Ultimately, it works because it gives you a rational parameterization of the unit circle.

Answer 2

Computer algebra systems can do this. Maybe this is not needed, since the question (answered by Steven) was only whether closed form exists (not to exhibit it). But just for fun I gave it to Maple...

$$ 1/5\,i\sqrt {\sqrt {5}+2}\ln \left( \left( \tan \left( 1/2\,x \right) \right) ^{2}+ \left( 1/2\,i \left( \sqrt {5}+2 \right) ^{3/2 }+1/2+1/2\,\sqrt {5}-5/2\,i\sqrt {\sqrt {5}+2} \right) \tan \left( 1/2 \,x \right) -1/2\,i \left( \sqrt {5}+2 \right) ^{3/2}+1/2-1/2\,\sqrt { 5}+5/2\,i\sqrt {\sqrt {5}+2} \right) \\-1/5\,i\sqrt {\sqrt {5}+2}\ln \left( \left( \tan \left( 1/2\,x \right) \right) ^{2}+ \left( -1/2 \,i \left( \sqrt {5}+2 \right) ^{3/2}+1/2+1/2\,\sqrt {5}+5/2\,i\sqrt { \sqrt {5}+2} \right) \tan \left( 1/2\,x \right) +1/2\,i \left( \sqrt { 5}+2 \right) ^{3/2}+1/2-1/2\,\sqrt {5}-5/2\,i\sqrt {\sqrt {5}+2} \right) \\+1/5\,\sqrt {-2+\sqrt {5}}\ln \left( \left( \tan \left( 1/2 \,x \right) \right) ^{2}+ \left( -1/2\, \left( -2+\sqrt {5} \right) ^ {3/2}+1/2-1/2\,\sqrt {5}-5/2\,\sqrt {-2+\sqrt {5}} \right) \tan \left( 1/2\,x \right) +1/2\, \left( -2+\sqrt {5} \right) ^{3/2}+1/2+1 /2\,\sqrt {5}+5/2\,\sqrt {-2+\sqrt {5}} \right) \\-1/5\,\sqrt {-2+\sqrt {5}}\ln \left( \left( \tan \left( 1/2\,x \right) \right) ^{2}+ \left( 1/2\, \left( -2+\sqrt {5} \right) ^{3/2}+1/2-1/2\,\sqrt {5}+5/ 2\,\sqrt {-2+\sqrt {5}} \right) \tan \left( 1/2\,x \right) -1/2\, \left( -2+\sqrt {5} \right) ^{3/2}+1/2+1/2\,\sqrt {5}-5/2\,\sqrt {-2+ \sqrt {5}} \right) \\+4/5\,\sqrt {2}{\rm{ arctanh}} \left( 1/4\, \left( 2 \,\tan \left( 1/2\,x \right) -2 \right) \sqrt {2} \right) $$ In this case, we only had to solve two successive quadratic equations.

Answer 3

Given $\displaystyle \int\frac{1}{\sin^5 x+\cos^5 x}dx$

First we will simplify $\sin^5 x+\cos^5 x = \left(\sin^2 x+\cos^2 x\right)\cdot \left(\sin^3 x+\cos^3 x\right) - \sin ^2x\cdot \cos ^2x\left(\sin x+\cos x\right)$

$\displaystyle \sin^5 x+\cos^5 x= (\sin x+\cos x)\cdot (1-\sin x\cdot \cos x-\cos^2 x\cdot \sin^2x)$

So Integral is $\displaystyle \int\frac{1}{\sin^5 x+\cos^5 x}dx $

$\displaystyle = \int\frac{1}{(\sin x+\cos x)\cdot (1-\sin x\cdot \cos x-\cos^2 x\cdot \sin^2x)}dx$

$\displaystyle = \int \frac{(\sin x+\cos x)}{(\sin x+\cos x)^2\cdot (1-\sin x\cdot \cos x-\cos^2 x\cdot \sin^2x)}dx$

$\displaystyle = \int \frac{(\sin x+\cos x)}{(1+\sin 2x)\cdot (1-\sin x\cdot \cos x-\cos^2 x\cdot \sin^2x)}dx$

Let $(\sin x-\cos x) = t\;,$ Then $(\cos +\sin x)dx = dt$ and $(1-\sin 2x) = t^2\Rightarrow (1+\sin 2x) = (2-t^2)$

So Integral Convert into $\displaystyle = 4\int\frac{1}{(2-t^2)\cdot(5-t^4)}dt = 4\int\frac{1}{(t^2-2)\cdot (t^2-\sqrt{5})\cdot (t^2+\sqrt{5})}dt$

Now Using partial fraction, we get

$\displaystyle = 4\int \left[\frac{1}{2-t^2}+\frac{1}{(2-\sqrt{5})\cdot 2\sqrt{5}\cdot (\sqrt{5}-t^2)}+\frac{1}{(2+\sqrt{5})\cdot 2\sqrt{5}\cdot (\sqrt{5}+t^2)}\right]dt$

$ = \displaystyle \sqrt{2}\ln \left|\frac{\sqrt{2}+t}{\sqrt{2}-t}\right|+\frac{1}{(2-\sqrt{5})\cdot 5^{\frac{3}{4}}}\cdot \ln \left|\frac{5^{\frac{1}{4}}+t}{5^{\frac{1}{4}}-t}\right|+\frac{2}{(2+\sqrt{5})\cdot 5^{\frac{3}{4}}}\cdot \tan^{-1}\left(\frac{t}{5^{\frac{1}{4}}}\right)+\mathbb{C}$

where $t=(\sin x-\cos x)$

Answer 4

Utilize the identity $$\sin^5 (t+\frac\pi4) +\cos^5(t+\frac\pi4) =\sqrt2 \cos t\left(\frac{5}4-\cos^4 t\right) $$ to integrate \begin{align} &\int \dfrac {1}{\sin^5 (t+\frac\pi4) +\cos^5(t+\frac\pi4)}dt\\ =& \frac{2\sqrt2}5 \int\bigg(\frac1{\cos^2 t}-\frac{1}{2\cos^2 t+\sqrt5} -\frac{1}{2\cos^2 t-\sqrt5} \bigg)\cos t \>dt \\ =& \frac25\bigg(\sqrt2\tanh^{-1}\sin t -\frac{\tanh^{-1}\frac{\sqrt2\sin t}{\sqrt{\sqrt5+2}}}{\sqrt{\sqrt5+2}} +\frac{\tan^{-1}\frac{\sqrt2 \sin t}{\sqrt{\sqrt5-2}}}{\sqrt{\sqrt5-2}}\bigg)+C \end{align} Then, replace $t=x-\frac\pi4$ to arrive at the original integral. In particular \begin{align} &\int_0^{\frac\pi2}\frac {dx}{\sin^5 x +\cos^5 x} =\int_{-\frac\pi4}^{\frac\pi4} \dfrac {1}{\sin^5 (t+\frac\pi4) +\cos^5(t+\frac\pi4)}dt \\ =&\frac45\bigg( \sqrt2\coth^{-1}\sqrt2-\frac{\coth^{-1} \sqrt{\sqrt5+2}}{\sqrt{\sqrt5+2}} + \frac{\cot^{-1} \sqrt{\sqrt5-2}}{\sqrt{\sqrt5-2}}\bigg) \end{align}