Math Problem Statement

Find the indefinite integral using the substitution x = 3 tan(θ). ∫ x √(9 + x²) dx

Solution

The problem asks to find the indefinite integral of the given function using the substitution x=3tan(θ)x = 3 \tan(\theta). Let's proceed step by step.

We are given the integral: x9+x2dx\int x \sqrt{9 + x^2} \, dx with the substitution x=3tan(θ)x = 3 \tan(\theta).

Step 1: Substitution

Since x=3tan(θ)x = 3 \tan(\theta), we differentiate to find dxdx: dx=3sec2(θ)dθdx = 3 \sec^2(\theta) \, d\theta

Next, substitute x=3tan(θ)x = 3 \tan(\theta) into 9+x2\sqrt{9 + x^2}: x2=9tan2(θ)x^2 = 9 \tan^2(\theta) Thus: 9+x2=9+9tan2(θ)=9(1+tan2(θ))=9sec2(θ)9 + x^2 = 9 + 9 \tan^2(\theta) = 9(1 + \tan^2(\theta)) = 9 \sec^2(\theta) So: 9+x2=3sec(θ)\sqrt{9 + x^2} = 3 \sec(\theta)

Step 2: Rewrite the integral in terms of θ\theta

Substitute x=3tan(θ)x = 3 \tan(\theta) and dx=3sec2(θ)dθdx = 3 \sec^2(\theta) \, d\theta into the integral: x9+x2dx=3tan(θ)3sec(θ)3sec2(θ)dθ\int x \sqrt{9 + x^2} \, dx = \int 3 \tan(\theta) \cdot 3 \sec(\theta) \cdot 3 \sec^2(\theta) \, d\theta This simplifies to: 27tan(θ)sec3(θ)dθ27 \int \tan(\theta) \sec^3(\theta) \, d\theta

Step 3: Solve the integral

We now need to solve: 27tan(θ)sec3(θ)dθ27 \int \tan(\theta) \sec^3(\theta) \, d\theta We use the identity tan(θ)=sin(θ)cos(θ)\tan(\theta) = \frac{\sin(\theta)}{\cos(\theta)} and express sec(θ)=1cos(θ)\sec(\theta) = \frac{1}{\cos(\theta)}, so the integral becomes: 27sin(θ)cos4(θ)dθ27 \int \frac{\sin(\theta)}{\cos^4(\theta)} \, d\theta We can solve this using substitution. Let u=cos(θ)u = \cos(\theta), so du=sin(θ)dθdu = -\sin(\theta) \, d\theta, and the integral becomes: 27duu4-27 \int \frac{du}{u^4} This is a standard integral: 27u4du=27(u33)=273u3=9u3-27 \int u^{-4} \, du = -27 \left( \frac{u^{-3}}{-3} \right) = \frac{27}{3} u^{-3} = 9 u^{-3} Substitute u=cos(θ)u = \cos(\theta) back in: 9sec3(θ)9 \sec^3(\theta)

Step 4: Final substitution back to xx

Since sec(θ)=9+x23\sec(\theta) = \frac{\sqrt{9 + x^2}}{3}, we have: 9sec3(θ)=9(9+x23)3=(9+x2)3/299 \sec^3(\theta) = 9 \left( \frac{\sqrt{9 + x^2}}{3} \right)^3 = \frac{(9 + x^2)^{3/2}}{9} Thus, the final answer is: (9+x2)3/29+C\frac{(9 + x^2)^{3/2}}{9} + C

Conclusion:

The indefinite integral is: x9+x2dx=(9+x2)3/29+C\int x \sqrt{9 + x^2} \, dx = \frac{(9 + x^2)^{3/2}}{9} + C

Do you need further details or clarifications? Here are 5 related questions to deepen understanding:

  1. How do you solve integrals involving trigonometric substitutions?
  2. What are the key trigonometric identities used in substitution methods?
  3. How do you reverse trigonometric substitution after solving an integral?
  4. Can you explain the relationship between tan(θ)\tan(\theta) and sec(θ)\sec(\theta) in solving integrals?
  5. What are the standard forms of integrals involving secn(θ)\sec^n(\theta)?

Tip: Always remember to revert back to the original variable after solving an integral using substitution!

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Math Problem Analysis

Mathematical Concepts

Integration
Trigonometric Substitution

Formulas

x = 3 tan(θ)
dx = 3 sec²(θ) dθ
∫ x √(9 + x²) dx
9 + x² = 9 sec²(θ)
√(9 + x²) = 3 sec(θ)
27 ∫ tan(θ) sec³(θ) dθ

Theorems

Trigonometric Substitution

Suitable Grade Level

Undergraduate Calculus