<s>[INST] Trigonometric integral I would like to know if there is some way to calculate exactly
$$\int_{0}^{\pi/2} \frac{\cos^3 x}{\sqrt {\sin x}(1+\sin^3 x)}\,dx$$
The numerical value is 1,52069
 [/INST] From Dilip Sarwate's suggestion on:
$$2\int_0^1\frac{1-y^4}{1+y^6}\mathrm dy = 2\int_0^1\frac{(1-y^2)(1+y^2)}{(1+y^2)(y^4-y^2+1)}\mathrm dy = 2\int_0^1\frac{(1-y^2)}{(y^4-y^2+1)}\mathrm dy \; .$$
The denominator can be split further into
$$2\int_0^1\frac{(1-y^2)}{(y^4-y^2+1)}\mathrm dy = 2\int_0^1\frac{(1-y^2)}{(y^2+\sqrt{3}y+1)(y^2-\sqrt{3}y+1)}\mathrm dy \; .$$
Using partial fractions this becomes
$$2\int_0^1\frac{(1-y^2)}{(y^2-\sqrt{3}y+1)(y^2+\sqrt{3}y+1)}\mathrm dy = 2\int_0^1\frac{\frac{y}{\sqrt{3}}+\frac{1}{2}}{(y^2+\sqrt{3}y+1)}-\frac{\frac{y}{\sqrt{3}}-\frac{1}{2}}{(y^2-\sqrt{3}y+1)}\mathrm dy \; .$$
The last part can be rewritten as
$$\frac{1}{\sqrt{3}}\int_0^1\frac{2y+\sqrt{3}}{(y^2+\sqrt{3}y+1)}-\frac{2y-\sqrt{3}}{(y^2-\sqrt{3}y+1)}\mathrm dy \; .$$
And can easily be solved by substitution to give as Zev Chonoles already mentioned in the commentary: $\frac{\ln(7+4\sqrt{3})}{\sqrt{3}}$ .
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