Calculate the integral: $\int\frac{\cos^5(t)}{(\sin(t))^{\frac{1}{3}}}dt$ I am in need of support from the community to help me in figuring out the calculation for the following integral:
$$\int\frac{\cos^5(t)}{(\sin(t))^{\frac{1}{3}}}dt$$
my approach:
$$\cos^5(t) = \cos(t)\left(\frac{1+\cos(2t)}{2}\right)^2 = \frac{1}{4}\bigl[\cos(t)(1+2\cos(2t)+\cos^2(2t))\bigr]$$
Then simplifying the integral:
$$\frac{1}{4}\left[\int \frac{\cos(t)}{(\sin(t))^{\frac{1}{3}}}+2\int\frac{\cos(2t)}{(\sin(t))^{\frac{1}{3}}}+\int\frac{\cos^2(2t)}{(\sin(t))^{\frac{1}{3}}} \right]dt$$
Using $u$ substitution for the first integral as $u = \sin(t)$ and $du = \cos(t)\,dt$
First integral: $\frac{1}{4}\int u^{-\frac{1}{3}}\,du = \frac{1}{4}\left[\frac{3u^{\frac{2}{3}}}{2} \right]$
I'm stuck on the second integral as I cannot make way on how to substitute to get rid of $t$and $2t$ respectively, any ideas?
 A: Substitue $u = \sin t \implies \mathrm{d}u = \cos t \mathrm{d}t$
$$ \int\frac{\cos^5(t)}{(\sin(t))^{\frac{1}{3}}}\mathrm{d}t = \int\frac{\cos^4(t)}{(\sin(t))^{\frac{1}{3}}} \cos t\mathrm{d}t = \int\frac{(1-u^2)^2}{u^{\frac{1}{3}}}\mathrm{d}u = \int (u^{-\frac13} -2u^{\frac53}+u^{\frac{11}{3}})\mathrm{d}u$$
A: Note that\begin{align}\cos^5(t)&=\cos(t)\cos^4(t)\\&=\cos(t)\bigl(1-\sin^2(t)\bigr)^2.\end{align}So, if you do $\sin(t)=x$ and $\cos(t)\,\mathrm dt=\mathrm dx$, then your integral becomes$$\int\frac{(1-x^2)^2}{\sqrt[3]x}\,\mathrm dx.$$Can you take it form here?
A: $$\int\frac{\cos^5(t)}{(\sin(t))^{\frac{1}{3}}}\,dt = \int\frac{\cos^4(t)\times \cos (t)}{(\sin(t))^{\frac{1}{3}}}\,dt=\int\frac{(\cos^2(t))^2\times \cos (t)}{(\sin(t))^{\frac{1}{3}}}\,dt=\int\frac{(1-\sin^2(t))^2\times \cos (t)}{(\sin(t))^{\frac{1}{3}}}\,dt=\int\frac{(1-u^2)^2}{u^{\frac{1}{3}}}\,du=\int\frac{1-2u^2+u^4}{u^{\frac{1}{3}}}\,du=\int ({u^{-\frac{1}{3}}-2u^{\frac{5}{3}}+u^{\frac{11}{3}})\,du}=\frac{3}{2}u^\frac{2}{3}-\frac{3}{4}u^\frac{8}{3}+ \frac{3}{14}u^{\frac{14}{3}}+c=\frac{3}{2}\sin^\frac{2}{3}t-\frac{3}{4}\sin^\frac{8}{3}t + \frac{3}{14}\sin^{\frac{14}{3}}t+c$$
$u=\sin t, du= \cos t \,dt$
