Apparent simple definite integral: $ \int_{0}^{1} x\sqrt{3-x^2-2x \cos a} \,dx $ This integral looks simple but I got stuck:
$$
\int_{0}^{1} x\sqrt{3-x^2-2x \cos a} \,dx 
$$
with $0 \leq a \leq 2 \pi$.
I tried to solve it by parts but it gets complicated. Also using some substitutions seems awkward.
Hints would be preferable than complete answer...
 A: Hint. One may perform the change of variable
$$
x+\cos a=\sin t \cdot \sqrt{3+\cos^2 a},\quad dx=\cos t \cdot \sqrt{3+\cos^2 a}\cdot dt,
$$ obtaining
$$
\int_{0}^{1} x\sqrt{3-x^2-2x \cos a} \,dx=(3+\cos^2 a)\int_{0}^{t_1} \left(\sin t\cdot\sqrt{3+\cos^2 a}-\cos a \right)\cos^2 t\cdot dt
$$ which is a classic trigonometric integral easy to evaluate, with
$$
t_1=\arcsin \frac{1+\cos a}{\sqrt{3+\cos^2 a}}.
$$
A: Write
$$\int_0^1 x\sqrt{3-x^2-2x\cos a}\, dx $$
$$= -\frac 1 2\int_0^1 (-2x-2\cos a)\sqrt{3-x^2-2x\cos a} - \cos a\int_0^1 \sqrt{3-x^2-2x\cos a}\, dx$$
The first integral can be solved with $u$-substitution. The second is harder. Before reading on, here's a hint: try getting that second integral into a form that looks like $\int_0^1 \sqrt{1-u^2}\, du$, and use trigonometric substitution. 
Here's a sketch on how to complete the second integral. Complete the square under the radical to get it into the form of $b^2+(cx+d)^2$, and then let $u=\frac{cx+d}{b}$. Then the integral becomes
$$\int_{d/b}^{(c+d)/b}b\sqrt{1-u^2}\, du$$
This integral can be solved by letting $ = \sin \theta, du = \cos \theta\, d\theta$. 
