We are tasked with evaluating the integral:

\[
I = \int \frac{-4}{x^2 \sqrt{16x^2 - 9}} \, dx
\]

To solve this, we will use a substitution method. Let's follow the steps:

### Step 1: Simplifying the square root expression
We notice that \( \sqrt{16x^2 - 9} \) resembles a form that can benefit from a trigonometric substitution. Let’s try to simplify this expression first.

- Factor the term inside the square root:

\[
\sqrt{16x^2 - 9} = \sqrt{16\left(x^2 - \frac{9}{16}\right)}
\]

This leads us to consider a substitution of the form \( x = \frac{3}{4} \sec(\theta) \) because it matches the structure of \( a^2 \sec^2(\theta) - b^2 \). After applying this, we can proceed to solve the integral.

Let me calculate the result step by step.The integral evaluates to a piecewise result:

\[
I(x) = 
\begin{cases}
\frac{-4i \sqrt{-16 + \frac{9}{x^2}}}{9}, & \text{if} \ \frac{9}{x^2} > 16 \\
\frac{-4 \sqrt{16 - \frac{9}{x^2}}}{9}, & \text{otherwise}
\end{cases}
\]

This complex result suggests that the integral may not have a simple real solution over all values of \(x\) without considering conditions on the domain. Depending on the value of \(x\), the square root can result in complex values when the expression inside becomes negative.

Let me know if you would like further explanation of any steps!

### Follow-up Questions:
1. What conditions can ensure the integral remains real?
2. How would a trigonometric substitution simplify the square root?
3. What role does the imaginary unit \(i\) play in the result?
4. How would the answer change if the domain of \(x\) was restricted?
5. Can we interpret the physical significance of complex results in an applied problem?

**Tip**: Always check the domain of the function inside square roots when integrating to avoid complex results where real values are expected!

This article explains how to solve the integral ∫ (-4) / (x²√(16x² - 9)) using trigonometric substitution. The solution involves factoring, trigonometric substitution, and dealing with complex numbers when the expression inside the square root becomes negative. Suitable for advanced calculus students.

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