Question

Evaluate the integral of $x^3\sqrt{1 - x^2}$ with respect to $x$

Answer 1

$-\frac{1}{3}(1 - x^2)^{\frac{3}{2}} + \frac{1}{5}(1 - x^2)^{\frac{5}{2}} + C$

Answer 2

Solution (by substitution)

1. Let $u = 1 - x^2$.
2. Then $du = -2x\,dx$ ⟹ $x\,dx = -\tfrac{1}{2}\,du$.
3. Note $x^2 = 1 - u$.

Rewriting the integral:

$$\int x^3\sqrt{1 - x^2}\,dx = \int x^2 \cdot x \cdot u^{\tfrac12}\,dx = \int (1 - u)\,u^{\tfrac12}\,\bigl(x\,dx\bigr) = -\tfrac{1}{2}\int (1 - u)\,u^{\tfrac12}\,du.$$

Split and integrate termwise:

$$-\tfrac{1}{2}\int\bigl(u^{\tfrac12} - u^{\tfrac32}\bigr)\,du = -\tfrac{1}{2}\Bigl[\frac{2}{3}u^{\tfrac{3}{2}} - \frac{2}{5}u^{\tfrac{5}{2}}\Bigr] + C = -\tfrac{1}{3}u^{\tfrac{3}{2}} + \tfrac{1}{5}u^{\tfrac{5}{2}} + C.$$

Finally, substitute back $u = 1 - x^2$:

$$-\frac{1}{3}(1 - x^2)^{\frac{3}{2}} + \frac{1}{5}(1 - x^2)^{\frac{5}{2}} + C.$$