calculating $\int_0 ^3 x\sqrt{|1-x^2|}dx$ I tried to calculate $\int_0 ^3 x\sqrt{|1-x^2|}dx$ using $x=\sin t$, but couldn't solve it and got stuck. Is $x=\sin t$ correct or should I have used another value for $x$? don't know how to solve it.
hoping you can help me with that.
thank you very much for helping.

From x = 0 to x = 1, 
$$1 - {x^2} \ge 0$$
so
$$\left| {1 - {x^2}} \right| = 1 - {x^2}$$
From x = 1 to x = 3
$$\left| {1 - {x^2}} \right| \le 0$$
so 
$$\left| {1 - {x^2}} \right| = {x^2} - 1$$
$$\int_0^3 {x\sqrt {\left| {1 - {x^2}} \right|} dx}  = \int_0^1 {x\sqrt {1 - {x^2}} dx}  + \int_1^3 {x\sqrt {{x^2} - 1} } dx$$
Focusing on the first integral.. 
$$\int_0^1 {x\sqrt {1 - {x^2}} dx} $$
Let $$\boxed{u = 1-x^2}$$. It follows that:
$$\frac{{du}}{{dx}} = \frac{d}{{dx}}\left( {1 - {x^2}} \right) =  - 2x$$
$$du =  - 2xdx$$
$$xdx =  - \frac{{du}}{2}$$
$$x = 0 \to u = 1 - {0^2} = 1$$
$$x = 1 \to u = 1 - {1^2} = 0$$
\begin{equation}
\int_{x = 0}^{x = 1} {x\sqrt {1 - {x^2}} dx = }  - \frac{1}{2}\int_{u = 1}^{u = 0} {\sqrt u du}  =  - \frac{1}{2}\int_{u = 1}^{u = 0} {{u^{1/2}}du}
\end{equation}
If we switch the limits on the far right integral, that cancels out the $-$ sign...
$$\boxed{\int_{x = 0}^{x = 1} {x\sqrt {1 - {x^2}} dx = } \frac{1}{2}\int_{u = 0}^{u = 1} {\sqrt u du}}$$
The second integral... Let
$$\boxed{v = x^2 - 1}$$
It follows that:
$$\frac{{dv}}{{dx}} = \frac{d}{{dx}}\left( {{x^2} - 1} \right) = 2x$$
$$dv = 2xdx$$
$$xdx = \frac{{dv}}{2}$$
$$x = 1 \to v = {1^2} - 1 = 0$$
$$x = 3 \to v = {3^2} - 1 = 8$$
$$\boxed{\int_{x = 1}^{x = 3} {x\sqrt {{x^2} - 1} dx = } \frac{1}{2}\int_{v = 0}^{v = 8} {\sqrt v dv}  = \frac{1}{2}\int_{v = 0}^{v = 8} {{v^{1/2}}dv}}$$
So from the boxed equations above... we have
$$\boxed{\int_{x = 0}^{x = 3} {x\sqrt {\left| {1 - {x^2}} \right|} dx = }   \frac{1}{2}\int_{u = 0}^{u = 1} {\sqrt u du}  + \frac{1}{2}\int_{v = 0}^{v = 8} {\sqrt v dv}}$$ 
Do you know how to finish from there ^^ ?
