How do you integrate $\frac{\cos^5(x)}{\sin^5(x) + \cos^5(x)}$? 
$$\int_{0}^{\frac{\pi}{2}} \frac{\cos^5(x)}{\sin^5(x) + \cos^5(x)} \,dx$$

I tried by dividing the terms in both the numerator and denominator by $\cos^5x$ but still cant find my way.
 A: $$I =\int_{0}^{\frac{\pi}{2}} \frac{\cos^5x}{\sin^5x + \cos^5x} dx\tag{1}$$
Let $u =\frac{\pi}{2}-x$
$$I = -\int_{\frac{\pi}{2}}^{0} \frac{\sin^{5}u}{\sin^5u+\cos^5u}du\tag{2} =\int_{0}^{\frac{\pi}{2}} \frac{\sin^5x}{\sin^5x + \cos^5x} dx$$
Then $$(1)+(2)\implies 2I=\int_{0}^{\frac{\pi}{2}}\frac{\sin^5x+\cos^5x}{\sin^5x+\cos^5x}dx=\int_{0}^{\frac{\pi}{2}}1dx=\frac{\pi}{2}\implies I=\frac{\pi}{4}$$
A: The problem was changed and my following work is not necessary. 
Let $\tan{x}=t$.
Hence, $dt=\frac{1}{\cos^2x}dx=(1+t^2)dx$.
Thus,
$$\int\frac{\cos^5x}{\sin^5x+\cos^5x}dx=\int\frac{1}{1+\tan^5x}dx=\int\frac{1}{(1+t^2)(1+t^5)}dt.$$
Let $t+\frac{1}{t}=u$.
Hence, $$1+t^5=(1+t)(1-t+t^2-t^3+t^4)=t^2(1+t)\left(t^2+\frac{1}{t^2}-t-\frac{1}{t}+1\right)=$$
$$=t^2(1+t)(u^2-2-u-1)=t^2(1+t)(u^2-u-1)=t^2(1+t)\left(u-\frac{1-\sqrt5}{2}\right)\left(u-\frac{1+\sqrt5}{2}\right)=$$
$$=(1+t)\left(t^2-\frac{1-\sqrt5}{2}t+1\right)\left(t^2-\frac{1+\sqrt5}{2}t+1\right).$$
The rest is smooth:
Let $$\frac{1}{(1+t)(1+t^2)(\left(t^2-\frac{1-\sqrt5}{2}t+1\right)\left(t^2-\frac{1+\sqrt5}{2}t+1\right)}=$$
$$=\frac{A}{1+t}+\frac{Bt+C}{1+t^2}+\frac{Dt+E}{t^2-\frac{1-\sqrt5}{2}t+1}+\frac{Ft+G}{t^2-\frac{1+\sqrt5}{2}t+1}$$
and solve the system. 
