Which method to use to integrate this function? Recently I've been working on problems including integration and I'm having some difficulties solving this integral :
$$\int_0^1 (1-x)\sqrt{4x-x^2} {dx}$$
I tried to rewrite it in this form:
$$\int_0^1 (1-x)x\sqrt{\frac{4-x}{x}} dx$$ and then subsitute $\frac{4-x}{x}=t^2$ and by this substitution I get a function way easier integrated of this form:
$$-32\int\frac{(t^2-3)t^2}{(1+t^2)^4}dt$$
I can integrate this function easily by simplifying it but the problem is that I cannot define its borders because $t$ is not defined for $x=0$ so this means that my substitution doesn't hold. But I don't have any other idea how to solve it so any help will be appreciated.Thank you!
 A: One can use $\,\displaystyle{\int_a^b f(x) dx=\int_a^b f(a+b-x) dx}\,$ to get: $$I=\int_0^1 x \sqrt{4-(1+x)^2} dx= \int_0^1 (x+1)\sqrt{4 - (x+1)^2} dx-\int_0^1 \sqrt{4 - (x+1)^2} dx$$ For the first one just substitute $4-(x+1)^2 =t$ to get $\displaystyle{\frac12 \int_0^3 \sqrt t dt}.\, $ And the second one is a standard square root integral. 
A: Hint:
By completing the square,
$$4x-x^2=4-(x-2)^2$$ and that hints to use the substitution
$$x-2=2\cos t.$$
Then
$$\int (1-x)\sqrt{4x-x^2}dx=4\int(1+2\cos t)\sin^2t\,dt.$$
The term $\sin^2t$ is integrated in the form $\dfrac{1-\cos 2t}2$, and the second term is immediate.

 $$I=2t-2\sin t\cos t+\dfrac83\sin^3t=\\\arccos\dfrac{x-2}2-(x-2)\sqrt{1-\dfrac{(x-2)^2}4}+\dfrac83\left(1-\dfrac{(x-2)^2}4\right)^{3/2}$$

A: Your approach can be completed by noting that the domain of integrtion $x\in(0,1)$ gets transformed to the unbounded domain of integration $t\in(\sqrt3,\infty)$ with reversal of orientation (which negates the integral).

We can also set $u=1-x/2\implies4\!\left(1-u^2\right)=4x-x^2$. Therefore,
$$
\int_0^1(1-x)\sqrt{4x-x^2}\,\mathrm{d}x=4\int_{1/2}^1(2u-1)\sqrt{1-u^2}\,\mathrm{d}u
$$
Then, setting $u=\sin(\theta)$ gives
$$
\begin{align}
4\int_{1/2}^1(2u-1)\sqrt{1-u^2}\,\mathrm{d}u
&=4\int_{\pi/6}^{\pi/2}(2\sin(\theta)-1)\overbrace{\cos^2(\theta)}^{1-\sin^2(\theta)}\,\mathrm{d}\theta\\
&=4\int_{\pi/6}^{\pi/2}\left(-2\color{#C00}{\sin^3(\theta)}+\color{#090}{\sin^2(\theta)}+2\sin(\theta)-1\right)\mathrm{d}\theta\\
&=4\int_{\pi/6}^{\pi/2}\left(-2\color{#C00}{\frac{3\sin(\theta)-\sin(3\theta)}4}+\color{#090}{\frac{1-\cos(2\theta)}2}+2\sin(\theta)-1\right)\mathrm{d}\theta\\
%&=-\left[\frac23\cos(3\theta)+\sin(2\theta)+2\cos(\theta)+2\theta\right]_{\pi/6}^{\pi/2}\\[3pt]
%&=\frac32\sqrt3-\frac23\pi
\end{align}
$$
A: The us try to get rid of the square root step-by-step:
$$\begin{eqnarray*} \int_{0}^{1}(1-x)\sqrt{x}\sqrt{4-x}\,dx&\stackrel{x\mapsto z^2}{=}& 2\int_{0}^{1}z^2(1-z^2)\sqrt{4-z^2}\,dz\\&\stackrel{z\mapsto 2u}{=}&32\int_{0}^{1/2}u^2(1-4u^2)\sqrt{1-u^2}\,du\\&\stackrel{u\mapsto\sin\theta}{=}&32\int_{0}^{\pi/6}\sin^2\theta\cos^2\theta(1-4\sin^2\theta)\,d\theta\\&=&4\int_{0}^{\pi/6}\left[-1+\cos(2\theta)+\cos(4\theta)-\cos(6\theta)\right]\,d\theta.\end{eqnarray*}$$
I guess you may take it from here.
A: Hint: $(1-x)\sqrt{4x-x^2} {dx}=-\sqrt{4x-x^2}{dx} +(2-x)\sqrt{4x-x^2}{dx}=-\sqrt{4x-x^2}{dx} +0.5\sqrt{4x-x^2}{d(4x-x^2)}$
