Evaluate the integral $\int x^{\frac{-4}{3}}(-x^{\frac{2}{3}}+1)^{\frac{1}{2}}\mathrm dx$ $$x^{\frac{-4}{3}}(-x^{\frac{2}{3}}+1)^{\frac{1}{2}}=\frac{\sqrt{(-\sqrt[3]{x^2}+1)}}{\sqrt[3]{x^4}}$$
Is it necessary to simplify the function further? What substitution is useful?
$u=\sqrt[n]{\frac{ax+b}{cx+d}}$ doesn't work.
 A: As suggested in RecklessReckoner's comment, let us first change variable $x=u^3$ then $dx=3u^2du$. So, $$I=\int \frac{\sqrt{1-x^{2/3}}}{x^{4/3}}dx=3\int\frac{ \sqrt{1-u^2}}{u^2}du$$ Now, $u=\sin(t)$, $du=\cos(t)\,dt$ makes $$I=3\int  \cot ^2(t)\,dt=3\int \frac{1-\sin^2(t)}{\sin^2(t)}dt =3\Big(\int \frac{dt}{\sin^2(t)}-\int dt \Big)=-3 \big(t+\cot (t)\big)$$ For sure, we could have saved a step with a single change of variable $x=\sin^3(t)$.
A: Notice, $$\int x^{-4/3}\left(-x^{2/3}+1\right)^{1/2}\ dx$$$$=\int \frac{1}{x}\left(-x^{2/3}+1\right)^{1/2}(x^{-1/3}\ dx)$$
Let $-x^{2/3}+1=\sin^2\theta\implies -\frac{2}{3}x^{-1/3}\ dx=2\sin\theta\cos\theta\ d\theta$ or $x^{-1/3}\ dx=-3\sin\theta\cos\theta\ d\theta $
$$=\int\frac{1}{(1-\sin^2\theta)^{3/2}}(\sin\theta)(-3\sin\theta\cos\theta\ d\theta )$$
$$=-3\int\frac{\sin^2\theta\cos\theta}{\cos^3\theta}\ d\theta$$
$$=-3\int\tan^2\theta\ d\theta$$
$$=-3\int(\sec^2\theta-1)\ d\theta$$
$$=-3(\tan\theta-\theta)+C$$
$$=3\theta-3\tan\theta+C$$
substituting back the value of $\theta$, 
$$=\color{red}{3\sin^{-1}(\sqrt{1-x^{2/3}}) \ - \ 3 \frac{\sqrt{x^{-2/3}-1}}{x^{1/3}} \ + \ C}$$
A: Following RecklessReckoner's comment:
We have $\int x^{\frac{-4}{3}}(-x^{\frac{2}{3}}+1)^{\frac{1}{2}}\mathrm dx$.
Now, let $u=x^{1/3}$ and $3u \mathrm du=\mathrm dx$
This gives $3\int u^{-4}(-u^{2}+1)^{\frac{1}{2}}u\mathrm du=3\int u^{-3}(-u^{2}+1)^{\frac{1}{2}}\mathrm du=3\int u^{-3}(1-u^{2})^{\frac{1}{2}}\mathrm du.$
Next, we perform another substitution: let $u=\sin t$ and $\mathrm du=\cos t \mathrm dt$
$3\int u^{-3}(1-u^{2})^{\frac{1}{2}}\mathrm du=3\int {\sin^{-3} t}(1-{\sin}^{2} t)^{\frac{1}{2}}\cos t\mathrm dt=3\int {\sin^{-3} t}({\cos}^{2} t)^{\frac{1}{2}}\cos t\mathrm dt$
$=3\int {\sin^{-3} t} \,{\cos}^{2} t\mathrm dt=3\int \frac{\cos^2t}{\sin^3 t} \mathrm dt=3\int {\cot^2t} \,{\csc t} \mathrm dt$
We can use the identity $\csc^2 t-1= \cot^2 t.$
$3\int {\cot^2t} \,{\csc t} \mathrm dt=3\int (\csc^2 t-1)\csc t \mathrm dt=3\int \csc^3 t-\csc t  \mathrm dt=3\int \csc^3 t \mathrm dt- \int \csc t  \mathrm dt.$
For the second term use the fact that $\int \csc t \mathrm dt =-\log|\csc t+\cot t| \mathrm dt+C$
For the first term, use the reduction formula $$\int \csc^m t \mathrm dt= \frac{(-\cos t)(\csc^{m-1} t)}{m-1}+\frac{m-2}{m-1}\int \csc^{m-2} t \mathrm dt$$
with m=3.
Then, substitute back for $t$ and $u$. Can you take it from here? 
