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\( \int \frac { x ^ { 3 } d x } { \sqrt { ( r a - x ^ { 2 } ) ^ { 3 } } } \)

Ask by Pritchard Mathis. in Iran
Jan 12,2025

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The integral simplifies to: \[ \frac{ra}{\sqrt{ra - x^{2}}} + \sqrt{ra - x^{2}} + C \]

Solución

To evaluate the integral: \[ \int \frac{x^3 \, dx}{\sqrt{(ra - x^2)^3}} \] we can use substitution to simplify the expression. **Step 1: Substitution** Let: \[ u = ra - x^2 \] Then, the differential \( du \) is: \[ du = -2x \, dx \quad \Rightarrow \quad x \, dx = -\frac{du}{2} \] **Step 2: Rewrite the Integral** Express \( x^3 \, dx \) in terms of \( u \): \[ x^3 \, dx = x^2 \cdot x \, dx = (ra - u) \cdot \left(-\frac{du}{2}\right) = -\frac{(ra - u)}{2} \, du \] Substitute into the integral: \[ \int \frac{x^3 \, dx}{\sqrt{(ra - x^2)^3}} = \int \frac{-\frac{(ra - u)}{2} \, du}{u^{3/2}} = -\frac{1}{2} \int (ra - u) u^{-3/2} \, du \] **Step 3: Simplify and Integrate** Distribute the integrand: \[ -\frac{1}{2} \int ra \cdot u^{-3/2} \, du + \frac{1}{2} \int u^{-1/2} \, du \] Integrate each term: \[ - \frac{ra}{2} \cdot \left( -2 u^{-1/2} \right) + \frac{1}{2} \cdot \left( 2 u^{1/2} \right) + C = \frac{ra}{\sqrt{u}} + \sqrt{u} + C \] **Step 4: Substitute Back** Recall that \( u = ra - x^2 \): \[ \frac{ra}{\sqrt{ra - x^2}} + \sqrt{ra - x^2} + C \] Thus, the final answer is: \[ \boxed{ \frac{ra}{\sqrt{ra - x^{2}}} + \sqrt{ra - x^{2}} + C } \]

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Bonus Knowledge

To solve the integral \( \int \frac { x ^ { 3 } d x } { \sqrt { ( r a - x ^ { 2 } ) ^ { 3 } } } \), a common approach is to use a trigonometric substitution. You can substitute \( x = r \sin(\theta) \), which simplifies the expression. When you differentiate \( x \) (i.e., \( dx = r \cos(\theta)d\theta \)), the integral becomes more manageable because you can express the limits in terms of \( \theta \), leading to a simpler evaluation. After performing the substitution and simplifying, the integral can be resolved, often leading to results involving inverse trigonometric functions or expressions related to elliptic integrals depending on the bounds and constants involved. Now, let’s consider the wonderfully rich history of calculus! The integral you have here connects back to the works of mathematicians like Isaac Newton and Gottfried Wilhelm Leibniz, who independently developed the fundamental principles of calculus in the late 17th century. They opened the door to exploring areas under curves and solving complex problems with elegant techniques that transformed mathematics into the analytical framework we use today! For practical applications, integrals like this one often appear in physics, particularly in calculating moments of inertia and center of mass in mechanics. Engineers use such techniques in structural analysis and fluid dynamics where integration determines quantities related to distributions and flow behavior, making this integral more than just an academic exercise—it's a tool that helps us understand and design the world around us!

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