To evaluate the integral

$\int \frac{1}{(16x^2 - 1296)^{\frac{3}{2}}} \, dx$

for $x > 9$ using a trigonometric substitution, let's proceed as follows:

Step-by-Step Solution

1. Rewrite the Expression: Start by factoring out the constant under the square root to simplify the expression. Notice that: $16x^2 - 1296 = 16(x^2 - 81) = 16(x^2 - 9^2).$

2. Choose the Substitution: Since we have $x^2 - 9^2$, this suggests a trigonometric substitution with $x = 9 \sec \theta$, where $x > 9$ corresponds to $\sec \theta \geq 1$ (or $\theta$ in the range $[0, \pi/2)$).

3. Differentiate: With $x = 9 \sec \theta$, we find that $dx = 9 \sec \theta \tan \theta \, d\theta.$

4. Substitute into the Integral: Substitute $x = 9 \sec \theta$ and $dx = 9 \sec \theta \tan \theta \, d\theta$ into the integral: $\int \frac{1}{(16x^2 - 1296)^{\frac{3}{2}}} \, dx = \int \frac{1}{(16 \cdot 81 \sec^2 \theta - 1296)^{\frac{3}{2}}} \cdot 9 \sec \theta \tan \theta \, d\theta.$

5. Simplify the Expression Inside the Integral: Simplify the expression in the denominator: $16 \cdot 81 \sec^2 \theta - 1296 = 1296 (\sec^2 \theta - 1) = 1296 \tan^2 \theta.$ Thus, $(16x^2 - 1296)^{\frac{3}{2}} = (1296 \tan^2 \theta)^{\frac{3}{2}} = 1296^{\frac{3}{2}} \tan^3 \theta = 46656 \tan^3 \theta.$

6. Rewrite the Integral: Substitute back into the integral: $\int \frac{1}{46656 \tan^3 \theta} \cdot 9 \sec \theta \tan \theta \, d\theta = \frac{9}{46656} \int \frac{\sec \theta}{\tan^2 \theta} \, d\theta.$

7. Simplify Further: Rewrite $\frac{\sec \theta}{\tan^2 \theta}$ as $\frac{\cos \theta}{\sin^2 \theta} = \csc^2 \theta \cot \theta$: $= \frac{9}{46656} \int \csc^2 \theta \, d\theta.$

8. Integrate: The integral of $\csc^2 \theta$ is $-\cot \theta$, so we have: $\frac{9}{46656} \cdot (-\cot \theta) = -\frac{9}{46656} \cot \theta.$

9. Back-Substitute: Since $x = 9 \sec \theta$, we have $\sec \theta = \frac{x}{9}$ and $\cot \theta = \frac{\sqrt{x^2 - 81}}{9}$. Substitute this back to obtain the result: $-\frac{9}{46656} \cdot \frac{\sqrt{x^2 - 81}}{9} = -\frac{\sqrt{x^2 - 81}}{46656}.$

Final Answer

$\int \frac{1}{(16x^2 - 1296)^{\frac{3}{2}}} \, dx = -\frac{\sqrt{x^2 - 81}}{46656} + C.$

Would you like additional steps or a deeper explanation of any part?

Related Questions

1. How would you solve a similar integral with a different constant in place of 1296?
2. Can trigonometric substitution be used for integrals involving $x^2 + a^2$?
3. What are some alternative methods for solving integrals of this type?
4. How does trigonometric substitution compare to hyperbolic substitution?
5. What is the importance of choosing the correct trigonometric identity in substitutions?

Tip

When performing trigonometric substitution, always simplify the expression first, as it often makes the substitution and integration process much easier.