# 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.

Hint:\begin{align}\int\frac{\mathrm dx}{\sin\left(\frac x2\right)\sqrt{\cos^3\left(\frac x2\right)}}&=\int\frac{\sin\left(\frac x2\right)}{\left(1-\cos^2\left(\frac x2\right)\right)\sqrt{\cos^3\left(\frac x2\right)}}\,\mathrm dx\\&=-2\int\frac{\mathrm dt}{(1-t^2)t\sqrt t}.\end{align}

• Tried it that way too. Didn't prove much beneficial to me – Rohan Shinde May 13 '18 at 14:50
• @Manthanein Now, doing $t=u^2$ and $\mathrm dt=2u\,\mathrm du$ turns this into the computation of a primitive of a rational function. – José Carlos Santos May 13 '18 at 14:56
• Oh yeah got that – Rohan Shinde May 13 '18 at 15:00

By performing the substitution $t=u^2$,
$$\int\frac{dt}{(1-t^2)t\sqrt t}=\int\frac{2udu}{(1-u^4)(u^3)}=2\int\frac{du}{(1-u^4)u^2}$$
$$\frac1{(1-u^4)u^2}$$ $$=\frac1{u^2}+\frac{u^2}{1-u^4}$$ $$=\frac1{u^2}+\frac{u^2}{2(1+u^2)}+\frac{u^2}{2(1-u^2)}$$ $$= \frac1{u^2}+\frac{u^2+1-1}{2(1+u^2)}+\frac{u^2-1+1}{2(1-u^2)}$$ $$=\frac1{u^2}+\frac12-\frac{1}{2(1+u^2)}-\frac12+\frac{1}{2(1-u^2)}$$ $$= \frac1{u^2}-\frac{1}{2(1+u^2)}+\frac{1}{2(1-u^2)}$$ $$= \frac1{u^2}-\frac{1}{2(1+u^2)}+\frac{1}{4(1-u)}+\frac{1}{4(1+u)}$$