Evaluate the integral using the indicated trigonometric substitution. (Use C for the constant of integration.)
x3

x2 + 49
 dx

,        x = 7 tan(𝜃)

To evaluate the integral \( \int \frac{x^3}{x^2 + 49} \, dx \) using the substitution \( x = 7 \tan(\theta) \), follow these steps:

### Step 1: Perform the trigonometric substitution
Given \( x = 7 \tan(\theta) \), we differentiate with respect to \( \theta \):
\[
dx = 7 \sec^2(\theta) \, d\theta
\]

Now substitute \( x = 7 \tan(\theta) \) into the original integral:
\[
x^2 + 49 = 49 \tan^2(\theta) + 49 = 49 (\tan^2(\theta) + 1) = 49 \sec^2(\theta)
\]

Also, \( x^3 = (7 \tan(\theta))^3 = 343 \tan^3(\theta) \).

Thus, the integral becomes:
\[
\int \frac{343 \tan^3(\theta)}{49 \sec^2(\theta)} \cdot 7 \sec^2(\theta) \, d\theta
\]
Simplifying:
\[
\int \frac{343 \tan^3(\theta) \cdot 7 \sec^2(\theta)}{49 \sec^2(\theta)} \, d\theta = \int \frac{343 \tan^3(\theta)}{7} \, d\theta
\]
\[
= 49 \int \tan^3(\theta) \, d\theta
\]

### Step 2: Simplify the integral
To simplify \( \int \tan^3(\theta) \, d\theta \), use the identity \( \tan^3(\theta) = \tan(\theta) \sec^2(\theta) - \tan(\theta) \). Thus, the integral becomes:
\[
49 \left( \int \tan(\theta) \sec^2(\theta) \, d\theta - \int \tan(\theta) \, d\theta \right)
\]

For \( \int \tan(\theta) \sec^2(\theta) \, d\theta \), perform the substitution \( u = \tan(\theta) \), so \( du = \sec^2(\theta) \, d\theta \). The integral becomes:
\[
\int u \, du = \frac{u^2}{2} = \frac{\tan^2(\theta)}{2}
\]

For \( \int \tan(\theta) \, d\theta \), the result is \( \ln|\sec(\theta)| \).

Thus, the integral becomes:
\[
49 \left( \frac{\tan^2(\theta)}{2} - \ln|\sec(\theta)| \right)
\]

### Step 3: Substitute back in terms of \( x \)
Recall that \( \tan(\theta) = \frac{x}{7} \) and \( \sec(\theta) = \sqrt{1 + \tan^2(\theta)} = \sqrt{1 + \frac{x^2}{49}} = \frac{\sqrt{x^2 + 49}}{7} \).

Substitute these back into the expression:
\[
49 \left( \frac{\left(\frac{x}{7}\right)^2}{2} - \ln\left|\frac{\sqrt{x^2 + 49}}{7}\right| \right)
\]
\[
= 49 \left( \frac{x^2}{98} - \ln\left|\frac{\sqrt{x^2 + 49}}{7}\right| \right)
\]
\[
= \frac{49x^2}{98} - 49 \ln\left|\frac{\sqrt{x^2 + 49}}{7}\right|
\]
\[
= \frac{x^2}{2} - 49 \ln\left|\frac{\sqrt{x^2 + 49}}{7}\right|
\]

### Step 4: Add the constant of integration
The final answer is:
\[
\boxed{\frac{x^2}{2} - 49 \ln\left|\frac{\sqrt{x^2 + 49}}{7}\right| + C}
\]

Do you have any questions or would like more details on a specific step?

---

Here are 5 related questions to further explore the topic:

1. How do you apply trigonometric substitution to integrals involving \( \sqrt{x^2 - a^2} \)?
2. What is the geometric interpretation of the substitution \( x = 7 \tan(\theta) \) in this context?
3. How do you derive the formula for integrating \( \tan^n(\theta) \) for general \( n \)?
4. What are the steps for simplifying integrals involving \( \sec(\theta) \) and \( \tan(\theta) \)?
5. How can you use hyperbolic functions for similar substitutions in integrals?

**Tip:** Always simplify the integral before substitution to make the final steps of the calculation easier.

Evaluate the integral using the indicated trigonometric substitution. (Use C for the constant of integration.) x^3 / (x^2 + 49) dx, x = 7 tan(𝜃)

Learn how to evaluate the integral of x^3 / (x^2 + 49) using the substitution x = 7 tan(𝜃). This detailed step-by-step solution covers trigonometric substitution and integration techniques, suitable for undergraduate calculus students.

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