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.
 A: First, we cannot substitute plain sine or cosine for $x$ because $x$ ranges up to $3$.  I suppose we could put $x = 3 \sin\theta$, but how does that help?  You still have positive and negative values in the absolute value brackets.  We should arrange to simplify those brackets out of our integral.
Since $1-x^2$ is nonnegative for $x \in [0,1]$ and nonpositive for $x \in [1,3]$, break the interval of integration so that you can discard the absolute value symbols.  \begin{align*}
  I &= \int_{0}^{3} \; x \sqrt{|1 - x^2|} \,\mathrm{d}x  \\
    &= \int_{0}^{1} \; x \sqrt{1 - x^2} \,\mathrm{d}x + \int_{1}^{3} \; x \sqrt{-(1 - x^2)} \,\mathrm{d}x
\end{align*}
Now, it sure would be handy if we had something simpler under those radicals.  Try substituting $u = 1 - x^2$ (with $\mathrm{d}u = -2x \,\mathrm{d}x$) in the first integral and $u = -(1 - x^2) = x^2 - 1$ (with $\mathrm{d}u = 2x \,\mathrm{d}x$) in the second.  (Happily, both these choices for $u$ are monotonic on their intervals, so we don't have to count preimages.)\begin{align*}
  I &= \int_{1-0^2 = 1}^{1-1^1 = 0} \; \frac{-1}{2} \sqrt{u} \,\mathrm{d}u + \int_{1^2 - 1 = 0}^{3^2 - 1 = 8} \; \frac{1}{2} \sqrt{u} \,\mathrm{d}u  \\
    &= \frac{1}{2} \int_{0}^{1} \; \sqrt{u} \,\mathrm{d}u + \frac{1}{2} \int_{0}^{8} \; \sqrt{u} \,\mathrm{d}u
\end{align*}
And you should be able to polish those off with the power rule.
A: Guide:
First split the region:
$$\int_0^3 x \sqrt{|1-x^2|}\, dx= \int_0^1 x \sqrt{1-x^2}\, dx+ \int_1^3 x \sqrt{x^2-1}\, dx$$
Evaluate the integral separately.
note that $\frac{d}{dx}x^2= 2x$.
A: $$\int_1^3 x \sqrt{x^2-1}\, dx$$
Apply substitution $u=\sqrt{x^2-1}$
$$\int_1^3 u^2\,du = \left[\frac{u^{2+1}}{2+1}\right]_1^3 =\left[\frac{\left(\sqrt{x^2-1}\right)^{2+1}}{2+1}\right]_1^3=\left[\frac{1}{3}\left(\sqrt{x^2-1}\right)^3\right]_1^3=\frac{16\sqrt{2}}{3}=7.8758\dots$$
A: 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 ^^ ?
