Math Problem Statement

Evaluate integral

Evaluate the integral \( \int \frac{dx}{x^2 \sqrt{25x^2 + 9}} \)

Solution

The integral provided is:

dxx225x2+9\int \frac{dx}{x^2 \sqrt{25x^2 + 9}}

To solve this, we will use a trigonometric substitution method. The form of the integral suggests that the substitution x=35tan(θ)x = \frac{3}{5} \tan(\theta) will simplify the square root term. Here's the step-by-step approach:

Step 1: Substitution

Let: x=35tan(θ)x = \frac{3}{5} \tan(\theta) Then: dx=35sec2(θ)dθdx = \frac{3}{5} \sec^2(\theta) d\theta

Now, substitute this into the square root term: 25x2+9=25(925tan2(θ))+9=9(tan2(θ)+1)=9sec2(θ)25x^2 + 9 = 25\left(\frac{9}{25}\tan^2(\theta)\right) + 9 = 9(\tan^2(\theta) + 1) = 9\sec^2(\theta) Thus: 25x2+9=3sec(θ)\sqrt{25x^2 + 9} = 3\sec(\theta)

Step 2: Substituting into the Integral

Substitute x=35tan(θ)x = \frac{3}{5} \tan(\theta) and dx=35sec2(θ)dθdx = \frac{3}{5} \sec^2(\theta) d\theta into the integral: dxx225x2+9=35sec2(θ)dθ(925tan2(θ))3sec(θ)\int \frac{dx}{x^2 \sqrt{25x^2 + 9}} = \int \frac{\frac{3}{5} \sec^2(\theta) d\theta}{\left(\frac{9}{25} \tan^2(\theta)\right) \cdot 3\sec(\theta)}

Simplifying the expression: =15sec2(θ)dθtan2(θ)sec(θ)= \frac{1}{5} \int \frac{\sec^2(\theta) d\theta}{\tan^2(\theta) \sec(\theta)} =15sec(θ)dθtan2(θ)= \frac{1}{5} \int \frac{\sec(\theta) d\theta}{\tan^2(\theta)}

Since sec(θ)=1cos(θ)\sec(\theta) = \frac{1}{\cos(\theta)} and tan(θ)=sin(θ)cos(θ)\tan(\theta) = \frac{\sin(\theta)}{\cos(\theta)}, we can further simplify the integral: =15cos(θ)dθsin2(θ)= \frac{1}{5} \int \frac{\cos(\theta) d\theta}{\sin^2(\theta)}

This is a standard integral, and it can be simplified as: 15csc2(θ)dθ=15cot(θ)\frac{1}{5} \int \csc^2(\theta) d\theta = -\frac{1}{5} \cot(\theta)

Step 3: Back-substitution

Now, we back-substitute θ\theta in terms of xx. Since x=35tan(θ)x = \frac{3}{5} \tan(\theta), we have: tan(θ)=5x3\tan(\theta) = \frac{5x}{3} Thus: cot(θ)=1tan(θ)=35x\cot(\theta) = \frac{1}{\tan(\theta)} = \frac{3}{5x}

Finally, the solution to the integral is: 1535x+C=325x+C-\frac{1}{5} \cdot \frac{3}{5x} + C = -\frac{3}{25x} + C

Final Answer:

dxx225x2+9=325x+C\int \frac{dx}{x^2 \sqrt{25x^2 + 9}} = -\frac{3}{25x} + C

Let me know if you need further clarification or details!


Here are 5 related questions that might interest you:

  1. How do you handle integrals involving other types of trigonometric substitutions?
  2. What are the key strategies for solving integrals with square roots of quadratic expressions?
  3. How can we apply partial fraction decomposition in integration?
  4. Can integrals involving hyperbolic functions simplify problems like this one?
  5. What is the geometric interpretation of trigonometric substitution in integrals?

Tip: Always look for patterns in the form of the integral, such as a2+x2a^2 + x^2, to choose the correct trigonometric substitution.

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Math Problem Analysis

Mathematical Concepts

Trigonometric Substitution
Definite Integrals
Integration

Formulas

Trigonometric substitution: x = (3/5) tan(θ)
dx = (3/5) sec²(θ) dθ
\int csc²(θ) dθ = -cot(θ)

Theorems

Trigonometric substitution theorem
Standard integral formulas

Suitable Grade Level

College Calculus