The appropriate trigonometric substitution is 
x = 7 sec(𝜃), which implies that dx = 7 sec(𝜃)tan(𝜃) d𝜃
. Next, write the integral in terms of 𝜃.


x2 − 49
x3
 dx



We are tasked with solving the integral:

\[
\int \frac{\sqrt{x^2 - 49}}{x^3} \, dx
\]

### Step 1: Apply the given trigonometric substitution
We are given that \( x = 7 \sec(\theta) \), which implies:

\[
dx = 7 \sec(\theta) \tan(\theta) \, d\theta
\]

Now, substitute \( x = 7 \sec(\theta) \) into the expression \( x^2 - 49 \):

\[
x^2 - 49 = (7 \sec(\theta))^2 - 49 = 49 \sec^2(\theta) - 49 = 49 (\sec^2(\theta) - 1)
\]

Using the identity \( \sec^2(\theta) - 1 = \tan^2(\theta) \), we get:

\[
x^2 - 49 = 49 \tan^2(\theta)
\]

Now we can rewrite the square root term:

\[
\sqrt{x^2 - 49} = \sqrt{49 \tan^2(\theta)} = 7 \tan(\theta)
\]

### Step 2: Substitute into the integral
Now substitute everything back into the original integral:

\[
\int \frac{\sqrt{x^2 - 49}}{x^3} \, dx = \int \frac{7 \tan(\theta)}{(7 \sec(\theta))^3} \cdot 7 \sec(\theta) \tan(\theta) \, d\theta
\]

Simplify the expression:

\[
= \int \frac{7 \tan(\theta)}{343 \sec^3(\theta)} \cdot 7 \sec(\theta) \tan(\theta) \, d\theta
= \int \frac{49 \tan^2(\theta)}{343 \sec^2(\theta)} \, d\theta
\]

Simplify further:

\[
= \frac{49}{343} \int \frac{\tan^2(\theta)}{\sec^2(\theta)} \, d\theta
= \frac{1}{7} \int \sin^2(\theta) \, d\theta
\]

### Step 3: Solve the integral
We now use the identity \( \sin^2(\theta) = \frac{1 - \cos(2\theta)}{2} \) to solve the integral:

\[
\frac{1}{7} \int \sin^2(\theta) \, d\theta = \frac{1}{7} \int \frac{1 - \cos(2\theta)}{2} \, d\theta
= \frac{1}{14} \int (1 - \cos(2\theta)) \, d\theta
\]

Now integrate:

\[
= \frac{1}{14} \left( \theta - \frac{\sin(2\theta)}{2} \right) + C
\]

### Step 4: Substitute back for \( \theta \)
Recall that \( x = 7 \sec(\theta) \), so \( \sec(\theta) = \frac{x}{7} \). Using \( \theta = \sec^{-1}\left(\frac{x}{7}\right) \), we substitute back in terms of \( x \).

Thus, the solution to the integral is:

\[
\frac{1}{14} \left( \sec^{-1}\left(\frac{x}{7}\right) - \frac{\sin(2\theta)}{2} \right) + C
\]

This completes the trigonometric substitution and integral evaluation.

Would you like to go into more detail, or do you have any questions?

---

Here are 5 related questions for practice:
1. How do you perform the substitution \( x = a \sin(\theta) \) in an integral?
2. What are the common trigonometric identities used in integration?
3. How do you evaluate integrals involving \( \sec(\theta) \) and \( \tan(\theta) \)?
4. How does the Pythagorean identity \( \sec^2(\theta) - \tan^2(\theta) = 1 \) help in solving integrals?
5. What is the process for converting a trigonometric integral back to the original variable?

**Tip:** Always double-check your trigonometric identities before solving integrals with trigonometric substitutions.

The appropriate trigonometric substitution is x = 7 sec(𝜃), which implies that dx = 7 sec(𝜃)tan(𝜃) d𝜃. Next, write the integral in terms of 𝜃.
x^2 − 49 / x^3 dx

This problem demonstrates the process of solving the integral (x^2 − 49) / x^3 dx using the trigonometric substitution x = 7 sec(θ). The solution includes detailed steps involving trigonometric identities, simplification, and final integration. Suitable for undergraduate calculus students.

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