# 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 Commented May 13, 2018 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. Commented May 13, 2018 at 14:56
• Oh yeah got that Commented May 13, 2018 at 15:00

By performing the substitution $t=u^2$,

$$\int\frac{dt}{(1-t^2)t\sqrt t}=\int\frac{2udu}{(1-u^4)(u^3)}=2\int\frac{du}{(1-u^4)u^2}$$

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.

Let $$y=\frac{1}{\sqrt{\cos \frac x2}},$$ then $$d y=\frac{1}{4} \sin \frac{x}{2 }\cdot\frac 1{\sqrt{\cos ^3 \frac{x}{2}}} d x$$, which transforms the integral into \begin{aligned} &\quad \int \frac{1}{\sin \frac{x}{2} \sqrt{\cos ^3 \frac{x}{2}}} d x \\&=\int \frac{4 d y}{\sqrt{1-\frac{1}{y^4}}} \\ &=4 \int \frac{y^2-1+1}{\sqrt{y^2-1}} d y \\ &=4 \int\left(\sqrt{y^2-1}+\int \frac{1}{\sqrt{y^2-1}}\right) d y \\ &=4\left[\frac { 1 } { 2 } \left(y \sqrt{y^2-1}-\ln \left(\sqrt{y^2-1}+y\right)\right)+\cosh ^{-1} y\right]+C\\&= 2\left[\sqrt{\sec \frac{x}{2}} \sqrt{\sec \frac{x}{2}-1}-\ln \left(\sqrt{\sec \frac{x}{2}-1}+\sqrt{\sec \frac{x}{2}}\right)\right]+ 4 \cosh ^{-1} \sqrt{\sec \frac{x}{2}}+C \end{aligned}

• I think that it should be $\frac{4 dy}{1-\frac{1}{y^4}}$ Commented Mar 23 at 12:48