<s>[INST] Evaluate $\int \frac {dx}{\sin \frac x2\sqrt {\cos^3 \frac x2}}$ 
Evaluate $$\int \frac {dx}{\sin \frac x2\sqrt {\cos^3 \frac x2}}$$

My try 
Write $t=\frac x2$ and hence $dx=2dt$
To change the integral to $$\int \frac {\csc t dt}{\cos^{\frac 32} t}$$
Multiplying both bottom and top by $\csc t$ and then using $\csc^2 t=1+\cot^2 t$ in the numerator the problem simplifies to $$2\int (\sin t)(\cos ^{\frac {-3}{2}} t) dt+2\int \frac {\cot^3 t dt}{\sqrt {\cos t}}$$
Now the first integral is easy to go but I am not getting any idea for the second one.  Any help would be very beneficial. New methods are also welcome. 
 [/INST] Elaborating on answer of @JoseCarlosSantos:
By performing the substitution $t=u^2$,
$$\int\frac{dt}{(1-t^2)t\sqrt t}=\int\frac{2udu}{(1-u^4)(u^3)}=2\int\frac{du}{(1-u^4)u^2}$$
Performing partial fraction decomposition,
$$\frac1{(1-u^4)u^2}$$
$$=\frac1{u^2}+\frac{u^2}{1-u^4}$$
$$=\frac1{u^2}+\frac{u^2}{2(1+u^2)}+\frac{u^2}{2(1-u^2)}$$
$$= \frac1{u^2}+\frac{u^2+1-1}{2(1+u^2)}+\frac{u^2-1+1}{2(1-u^2)}$$
$$=\frac1{u^2}+\frac12-\frac{1}{2(1+u^2)}-\frac12+\frac{1}{2(1-u^2)}$$
$$= \frac1{u^2}-\frac{1}{2(1+u^2)}+\frac{1}{2(1-u^2)} $$
$$= \frac1{u^2}-\frac{1}{2(1+u^2)}+\frac{1}{4(1-u)}+\frac{1}{4(1+u)} $$
which can be integrated easily.
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