use the substitution 5sin(u) = x. ∫x^2sqrt(25-x^2)dx

To evaluate the integral 

\[
\int x^2 \sqrt{25 - x^2} \, dx
\]

using the substitution \(5 \sin(u) = x\), follow these steps:

---

### Step 1: Substitution
From \(x = 5 \sin(u)\):
- \(dx = 5 \cos(u) \, du\)
- \(25 - x^2 = 25 - 25\sin^2(u) = 25\cos^2(u)\)

Substitute these into the integral:

\[
\int x^2 \sqrt{25 - x^2} \, dx = \int (5\sin(u))^2 \sqrt{25\cos^2(u)} (5\cos(u)) \, du
\]

---

### Step 2: Simplify the Expression
Simplify each term:
- \(x^2 = (5\sin(u))^2 = 25\sin^2(u)\)
- \(\sqrt{25 - x^2} = \sqrt{25\cos^2(u)} = 5\cos(u)\)

Thus, the integral becomes:

\[
\int x^2 \sqrt{25 - x^2} \, dx = \int 25\sin^2(u)(5\cos(u))(5\cos(u)) \, du
\]

Simplify further:

\[
\int x^2 \sqrt{25 - x^2} \, dx = 125 \int \sin^2(u)\cos^2(u) \, du
\]

---

### Step 3: Simplify Using Trigonometric Identities
Use the double-angle identity \(\sin^2(u)\cos^2(u) = \frac{1}{4}\sin^2(2u)\):

\[
125 \int \sin^2(u)\cos^2(u) \, du = 125 \int \frac{1}{4}\sin^2(2u) \, du = \frac{125}{4} \int \sin^2(2u) \, du
\]

---

### Step 4: Solve the Integral
Use the identity \(\sin^2(2u) = \frac{1}{2}(1 - \cos(4u))\) to rewrite the integral:

\[
\frac{125}{4} \int \sin^2(2u) \, du = \frac{125}{4} \int \frac{1}{2}(1 - \cos(4u)) \, du
\]

Simplify:

\[
\frac{125}{4} \int \frac{1}{2}(1 - \cos(4u)) \, du = \frac{125}{8} \int (1 - \cos(4u)) \, du
\]

Split the integral:

\[
\frac{125}{8} \int (1 - \cos(4u)) \, du = \frac{125}{8} \left(\int 1 \, du - \int \cos(4u) \, du\right)
\]

Evaluate each term:
- \(\int 1 \, du = u\)
- \(\int \cos(4u) \, du = \frac{\sin(4u)}{4}\)

Thus:

\[
\frac{125}{8} \left(u - \frac{\sin(4u)}{4}\right) = \frac{125}{8}u - \frac{125}{32}\sin(4u)
\]

---

### Step 5: Back-Substitute
Since \(x = 5\sin(u)\), \(\sin(u) = \frac{x}{5}\), and \(\cos(u) = \sqrt{1 - \sin^2(u)} = \sqrt{\frac{25 - x^2}{25}}\), we have:

- \(u = \arcsin\left(\frac{x}{5}\right)\)
- \(\sin(4u) = 4\sin(u)\cos(u)(1 - 2\sin^2(u)) = 4\frac{x}{5}\frac{\sqrt{25 - x^2}}{5}(1 - 2\frac{x^2}{25})\)

Substitute back to get the final expression.

---

Would you like detailed back-substitution steps or clarification on any part? 

Here are 5 questions related to this process for practice:

1. How do you verify that the substitution \(5\sin(u) = x\) simplifies the integral effectively?
2. What trigonometric identities are most useful in solving integrals of this form?
3. Can you perform a similar integral where \(x^2\sqrt{a^2 - x^2}\) has different coefficients?
4. How do you handle back-substitution for more complex trigonometric integrals?
5. How does symmetry in trigonometric functions affect definite integral evaluation?

**Tip:** Always simplify trigonometric integrals using standard identities before attempting to integrate!

Learn how to evaluate the integral ∫x²√(25 - x²)dx using the substitution 5sin(u) = x. This solution covers trigonometric substitution, simplifying integrals, and applying trigonometric identities for step-by-step results. Perfect for college-level calculus students.

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