solve the integral of square root of x^2 - 64 / x^3 dx using trig substituition 

To solve the integral \( \int \frac{\sqrt{x^2 - 64}}{x^3} \, dx \) using trigonometric substitution, follow these steps:

### Step 1: Substitution
Recognize that the expression \( \sqrt{x^2 - 64} \) suggests a trigonometric substitution of the form:

\[
x = 8 \sec(\theta)
\]

This substitution is chosen because it simplifies the square root term. Using this, we can compute the following:

\[
dx = 8 \sec(\theta) \tan(\theta) \, d\theta
\]

Now, let's rewrite the integral in terms of \( \theta \).

### Step 2: Simplify the Expression
First, substitute \( x = 8 \sec(\theta) \) into the square root part:

\[
\sqrt{x^2 - 64} = \sqrt{64 \sec^2(\theta) - 64} = \sqrt{64(\sec^2(\theta) - 1)} = \sqrt{64 \tan^2(\theta)} = 8 \tan(\theta)
\]

Now substitute into the entire integrand. Using \( x = 8 \sec(\theta) \), we get:

\[
\frac{\sqrt{x^2 - 64}}{x^3} = \frac{8 \tan(\theta)}{(8 \sec(\theta))^3} = \frac{8 \tan(\theta)}{512 \sec^3(\theta)} = \frac{1}{64} \cdot \frac{\tan(\theta)}{\sec^3(\theta)}
\]

Since \( \frac{\tan(\theta)}{\sec^3(\theta)} = \frac{\sin(\theta)}{\cos^4(\theta)} \), the integrand becomes:

\[
\frac{1}{64} \cdot \frac{\sin(\theta)}{\cos^4(\theta)}
\]

Thus, the integral is:

\[
\frac{1}{64} \int \frac{\sin(\theta)}{\cos^4(\theta)} \, d\theta
\]

### Step 3: Simplify the Integral
Now we can simplify the integrand using the identity \( \sin(\theta) = -d(\cos(\theta)) \), giving us:

\[
\int \frac{\sin(\theta)}{\cos^4(\theta)} \, d\theta = -\int \frac{d(\cos(\theta))}{\cos^4(\theta)} = -\int \cos^{-4}(\theta) \, d(\cos(\theta))
\]

Let \( u = \cos(\theta) \), then \( du = -\sin(\theta) \, d\theta \). The integral becomes:

\[
\frac{1}{64} \int u^{-4} \, du = \frac{1}{64} \cdot \frac{u^{-3}}{-3} = -\frac{1}{192} u^{-3}
\]

Substitute back \( u = \cos(\theta) \):

\[
-\frac{1}{192} \cdot \frac{1}{\cos^3(\theta)}
\]

### Step 4: Back-substitute in terms of \( x \)
Since \( \cos(\theta) = \frac{8}{x} \) (from \( x = 8 \sec(\theta) \)), we have:

\[
\cos^3(\theta) = \left(\frac{8}{x}\right)^3 = \frac{512}{x^3}
\]

Thus, the solution becomes:

\[
-\frac{1}{192} \cdot \frac{x^3}{512} = -\frac{x^3}{98304}
\]

### Step 5: Add the constant of integration
Finally, the integral is:

\[
-\frac{x^3}{98304} + C
\]

### Summary:
The solution to the integral is:

\[
\boxed{-\frac{x^3}{98304} + C}
\]

---

Would you like more details on any specific step or another type of explanation?

### Related Questions:
1. How does trigonometric substitution work for integrals involving square roots?
2. What are other forms of substitution for integrals involving \( x^2 - a^2 \)?
3. Can you explain how to solve integrals using hyperbolic substitution?
4. How do you handle integrals that involve higher powers of trigonometric functions?
5. What is the geometric interpretation of this type of substitution?

### Tip:
Always verify the derivative of your final answer to ensure that the original integrand is correctly obtained.

solve the integral of square root of x^2 - 64 / x^3 dx using trig substitution

This page provides a step-by-step solution to the integral of sqrt(x^2 - 64) / x^3 using trigonometric substitution. Learn key calculus concepts like substitution, simplification, and integration techniques. Suitable for undergraduate students.

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