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.

  • 2
    $\begingroup$ As long as the x's canceled you're good $\endgroup$ – randomgirl May 15 '18 at 0:58

$$\int x^3\sqrt{1-x^2}$$

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,

$$=-\frac12(-\int u^\frac32du+\int \sqrt{u}du)$$


Substituting back $u=1-x^2$


After doing small calculations,

$$\int x^3\sqrt{1-x^2}=-\frac12(-\frac25(1-x^2)^\frac52+\frac23(1-x^2)^\frac32)+C$$


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