- \(\displaystyle \int \frac {x^3}{\sqrt {25-x^2}} ~dx = \color {blue} {\frac {1}{3} (\sqrt {25-x^2})^3 - 25 \sqrt {25-x^2}}\)
Solution: \(\sin (u) = \frac {1}{5} x\), \(\cos (u) ~du = \frac {1}{5} ~dx\)
Solution: \(\frac {x^3}{\sqrt {25-x^2}} ~dx = \frac {\sin ^3 (u)}{125 \cos (u)} \cos (u) ~du = \frac {1}{125} \cdot \sin ^3 (u) ~du\)
Another Solution: \(u = \sqrt {25 - x^2}\), \(u ~du = - x ~dx\)
Another Solution: \(\frac {x^2}{\sqrt {25-x^2}} \cdot x ~dx = \frac {25 - u^2}{u} \cdot u ~du = (25 - u^2) ~du\)
Solution
Take \(|x| < 5\).
1. \(\int \frac{x^3}{\sqrt{25-x^2}} \, dx = \frac{1}{3}(25-x^2)^{3/2} - 25\sqrt{25-x^2} + C = \frac{1}{3}\sqrt{25-x^2}(-x^2 - 50) + C\).
1.1. Make the algebraic substitution \(u = \sqrt{25 - x^2}\).Then \(u^2 = 25 - x^2 \implies x^2 = 25 - u^2\), and \(2u \, du = -2x \, dx \implies x \, dx = -u \, du\).
1.2. Rewrite the integrand in terms of \(u\): \[\int \frac{x^3}{\sqrt{25-x^2}} \, dx = \int \frac{x^2 \cdot x \, dx}{\sqrt{25-x^2}} = \int \frac{25 - u^2}{u} (-u \, du) = \int (u^2 - 25) \, du.\]
1.3. Integrate polynomials in \(u\): \[\int (u^2 - 25) \, du = \frac{1}{3} u^3 - 25u + C.\]
1.4. Substitute back \(u = \sqrt{25-x^2}\): \[\frac{1}{3} (\sqrt{25-x^2})^3 - 25\sqrt{25-x^2} + C = \frac{1}{3}(25-x^2)^{3/2} - 25(25-x^2)^{1/2} + C.\]
2. The trigonometric substitution \(x = 5\sin(\theta)\) gives the same antiderivative.
2.1. Substitute \(x = 5\sin(\theta)\) with \(\theta \in (-\pi/2, \pi/2)\).Then \(dx = 5\cos(\theta)\, d\theta\) and \(\sqrt{25-x^2} = 5\cos(\theta)\).
2.2. The integral transforms into: \[\int \frac{125\sin^3(\theta)}{5\cos(\theta)} \cdot 5\cos(\theta) \, d\theta = 125 \int \sin^3(\theta) \, d\theta = 125 \int (1 - \cos^2(\theta)) \sin(\theta) \, d\theta.\]
2.3. Substituting \(w = \cos(\theta) \implies dw = -\sin(\theta)\, d\theta\): \[125 \int (w^2 - 1) \, dw = 125 \left(\frac{w^3}{3} - w\right) + C.\]
2.4. Since \(w = \cos(\theta) = \frac{\sqrt{25-x^2}}{5}\): \[125 \left(\frac{(\sqrt{25-x^2})^3}{3 \cdot 125} - \frac{\sqrt{25-x^2}}{5}\right) + C = \frac{1}{3}(\sqrt{25-x^2})^3 - 25\sqrt{25-x^2} + C.\]