Integrate $\left(\frac{\cos^3x\>+\>\sin^3x}{\cos^4x\>+\>\sin^4x}\right)^2$ over $[-\frac\pi4,\frac\pi4]$
I encountered the task of evaluate the definite integral, $$\int_{-\frac\pi4}^{\frac\pi4}\left(\frac{\cos^3x+\sin^3x}{\cos^4x+\sin^4x}\right)^2dx$$
which came about when I had answered a post as to whether there is an analytical solution to the area of an enclosed loop in polar space. After some laborious expansion and substitutions, I was happy to arrive at the algebraic expression $\frac{3\sqrt2}{8}\pi$, yet unhappy about the lengthy computation spent.
I wonder there may be some efficient procedure out there and would love to know.
Solution 1:
$$\int_{-\pi/4}^{\pi/4}\left(\frac{\sin^3 x+\cos^3 x}{\sin^4 x+\cos^4 x}\right)^2\,dx\stackrel{x\mapsto\arctan t}{=}\int_{-1}^{1}\frac{(1+t^3)^2}{(1+t^4)^2}\,dt\stackrel{\text{sym}}{=}2\int_{0}^{1}\frac{1+t^6}{(1+t^4)^2}\,dt $$
and the RHS (due to the sub $t\mapsto 1/t$) can also be expressed as
$$ \int_{0}^{+\infty}\frac{1+t^6}{(1+t^4)^2}\,dt = 2\int_{0}^{+\infty}\frac{dt}{(1+t^4)^2} $$
which only depends on values of the Beta function, after letting $\frac{1}{1+t^4}=u$.
By the same principle
$$ \int_{-\pi/4}^{\pi/4}\left(\frac{\sin^{2n-1}x+\cos^{2n-1}x}{\sin^{2n} x+\cos^{2n} x}\right)^2\,dx = \frac{\pi(2n-1)}{4n^2\sin\frac{\pi}{2n}} $$
for any $n>1$.
Solution 2:
If it's any help, the integrand may be simplified as follows, after setting $c=\cos x$ and $s=\sin x$:
In the denominator, we have that $$2(c^4+s^4)=2[(c^2)^2+(s^2)^2]=(c^2+s^2)^2+(c^2-s^2)^2=1+\cos^22x=1+\frac12(1+\cos 4x)=\frac32+\frac12\cos 4x.$$ Squaring gives $$\frac94+\frac32\cos4x+\frac14\cos^24x=\frac94+\frac32\cos4x+\frac18(1+\cos 8x)=\frac{19}{8}+\frac32\cos4x+\frac18\cos8x.$$
And now for the numerator, we have $$c^3+s^3=(c+s)(c^2+s^2-cs)=(c+s)(1-\frac12\sin 2x).$$ Squaring gives $$(c^2+s^2+2cs)(1-\sin 2x+\frac14\sin^22x)=(1+\sin 2x)[1-\sin 2x+\frac18(1-\cos 4x)]=(1+\sin 2x)(\frac98-\sin 2x-\frac18\cos 4x).$$ Expanding and simplifying gives $$\frac12+\frac{3}{16}\sin 2x-\frac{1}{16}\sin 6x+\frac38\cos 4x.$$
Thus your integrand becomes $$\frac{\frac12+\frac{3}{16}\sin 2x-\frac{1}{16}\sin 6x+\frac38\cos 4x}{\frac{19}{8}+\frac32\cos4x+\frac18\cos8x},$$ or more simply $$\frac{8+\sin 2x-\sin 6x+6\cos 4x}{38+24\cos4x+2\cos8x}.$$
It is either possible to notice a further simplification, or just use an appropriate substitution a la Weierstrass.