I'm trying to work out the following indefinite integral: $\int x^3 \sqrt{1-x^2}$. To solve this, I said let $u=1-x^2$. Then, $\frac{du}{dx}=-2x$, and $du=-2x \cdot dx$, and $dx=\frac{du}{-2x}$. Clearly, $x^3=-((1-x^2)-1)x=-(u-1)x$. I figure that this makes the integral: $\int (-u+1)x\sqrt{u}\cdot \frac{du}{-2x}=\int -\frac{(-u+1)\sqrt{u}}{2}\cdot du$
And I can solve it from there. Is that simplification process correct? I'm concerned that I can't have $x$ values when I'm integrating in terms of $u$, but they cancel out. Does that make it ok to include them?
Thanks for your time.
$$\int x^3\sqrt{1-x^2}$$
Apply u-substitution: $u=1-x^2$
$$=\int -\frac{(-u+1)\sqrt{u}}{2}du$$
$$=-\frac12\int(-u+1)\sqrt{u} du$$
$$=-\frac12\int -u^\frac{3}{2}+\sqrt{u}du$$
Apply the Sum Rule,
$$=-\frac12(-\int u^\frac32du+\int \sqrt{u}du)$$
$$=-\frac12(-\frac25u^\frac52+\frac23u^\frac32)$$
Substituting back $u=1-x^2$
$$=-\frac12(-\frac25(1-x^2)^\frac52+\frac23(1-x^2)^\frac32)$$
After doing small calculations,
$$\int x^3\sqrt{1-x^2}=-\frac12(-\frac25(1-x^2)^\frac52+\frac23(1-x^2)^\frac32)+C$$