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
 A: I am sure this substitution might work. Let $u=\sin x$ so that $du=\cos xdx$. Then your integrand becomes;
$$\int_0^{\pi/2}\frac{\cos^2x\cos xdx}{\sqrt \sin x(1+\sin^3x)}=\int_0^{\pi/2}\frac{(1-\sin^2x)\cos xdx}{\sqrt \sin x(1+\sin^3x)}=\int_0^1\frac{(1-u^2)du}{\sqrt u(1+u^3)}$$
$$=\int_0^1\frac{(1-u)(1+u)du}{\sqrt u(1+u)(1-u+u^2)}=\int_0^1\frac{(1-u)du}{\sqrt u(1-u+u^2)}$$
I believe at this point you can employ the method of partial fractions to integrate.
HINT: Let $u=v^2$ so that $du=2vdv$. Then you have $$\int_0^1\frac{(1-u)du}{\sqrt u(1-u+u^2)}=\int_0^1\frac{2(1-v^2)dv}{(1-v^2+v^4)}$$ and partial fraction follows. I think at this point the procedure for partial fraction is the same as above.
A: 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: $\frac{\ln(7+4\sqrt{3})}{\sqrt{3}}$ .
A: If I were in front of a classroom and a student asked me about this one, I'd begin by writing this:
$$\int_{0}^{\pi/2} \frac{\cos^3 x}{\sqrt {\sin x}(1+\sin^3 x)}\,dx = \int_{0}^{\pi/2} \frac{\cos^2 x}{\sqrt {\sin x}(1+\sin^3 x)}{\Huge(}\cos x \;dx{\Huge)}$$
If they don't know how to construe that as a hint, I'd tell them that's what they need to work on.
The point is that
$$
du = \cos x\; dx,
$$
so that suggests $u=\sin x$.
On the bottom we've got $\sqrt{u}(1+u^3)$.  On top we need to change $\cos^2 x$ to $1-\sin^2 x$ so that we'll have $1-u^2$.
A: After we deduce our integral is equal to $$I= 2\int_0^1\frac{(1-y^2)}{(y^4-y^2+1)}\mathrm dy$$ we can finish off like this: $$ I = -2 \int^1_0 \frac{ y^2 (1 - 1/y^2) }{ y^2 (y^2-1 + 1/y^2) } dy = -2 \lim_{a\to 0^+} \int^1_a \frac{ d(y+1/y) }{(y+1/y)^2-3} =2 \int_2^{\infty} \frac{du}{u^2-3}=\frac{\ln(7+4\sqrt{3})}{\sqrt{3}}.$$
