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 $$t=(\sin x-\cos x)$$
\begin{align}
\int\frac{\sqrt{\cos 2x}}{\sin x}\ dx&=\int\frac{\sqrt{\cos^2x-\sin^2x}}{\sin x}\ dx\\
&\stackrel{\color{red}{[1]}}=\int\frac{\sqrt{t^4-6t^2+1}}{t^3+t}\ dt\\
&\stackrel{\color{red}{[2]}}=\frac12\int\frac{\sqrt{u^2-6u+1}}{u^2+u}\ du\\
&\stackrel{\color{red}{[3]}}=\int\frac{(y^2-6y+1)^2}{(y-1)(y-3)(y+1)(y^2+2t-7)}\ dy\\
&\stackrel{\color{red}{[4]}}=\int\left[\frac1{y-1}+\frac1{y-3}-\frac1{y+1}-\frac{16}{y^2+2y-7}\right]\ dt\\
&=\int\left[\frac1{y-1}+\frac1{y-3}-\frac1{y+1}-\frac{16}{(y+1)^2-8}\right]\ dt
\end{align}
The rest is yours.
Notes :
$\color{red}{[1]}\;\;\;$Use Weierstrass substitution, $\tan\left(\dfrac{x}{2}\right)=t$.
$\color{red}{[2]}\;\;\;$Use substitution $u=t^2$.
$\color{red}{[3]}\;\;\;$Use Euler substitution, $y-u=\sqrt{u^2-6u+1}\;\color{blue}{\Rightarrow}\;y=\dfrac{u^2-1}{2u-6}$.
$\color{red}{[4]}\;\;\;$Use partial fractions decomposition.
Best Answer
If you do $u=\sin x$, then you must also do $\mathrm du=\cos x\,\mathrm dx$. So$$\int\frac{\cos x}{\sin^3x+\sin x}\,\mathrm dx$$becomes$$\int\frac1{u^3+u}\,\mathrm du=\int\frac1{u(u^2+1)}\,\mathrm du.$$Can you take it from here?