How to Solve the Indefinite Integral with Trigonometric Substitution?

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SUMMARY

The discussion focuses on solving the indefinite integral \(\int{ x^3 \sqrt{(36-x^2)}dx}\) using trigonometric substitution. The initial attempt involved substituting \(6\cos(\theta) = x\) and resulted in a complex expression that was deemed incorrect. A more effective approach suggested rewriting the integral as \(\int x^2\sqrt{36- x^2}(x dx)\) and using the substitution \(u = 36 - x^2\), leading to \(du = 2x dx\) and \(x^2 = 36 - u\). This method simplifies the integral significantly.

PREREQUISITES
  • Understanding of trigonometric substitution in integrals
  • Familiarity with integration techniques, particularly for polynomial and radical functions
  • Knowledge of basic calculus concepts, including derivatives and integrals
  • Ability to manipulate algebraic expressions and perform substitutions
NEXT STEPS
  • Study trigonometric substitution methods in calculus
  • Learn how to apply integration by parts for complex integrals
  • Explore the use of substitution in definite integrals
  • Practice solving integrals involving radicals and polynomials
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Students studying calculus, particularly those focusing on integration techniques, as well as educators looking for effective methods to teach trigonometric substitution.

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Homework Statement




\int{ x^3 \sqrt{(36-x^2)}dx}

Homework Equations





The Attempt at a Solution



I tried using trig substitution but got 7776\int{cos^3(\theta)-cos^5(\theta)d\theta} which seems completely wrong



6cos(\theta)=x <br />
6sin(\theta)d\theta=dx <br />
6sin(\theta)=\sqrt{36-x^2} <br />

\int{ (6cos(\theta))^36sin(\theta)6sin(\theta)d\theta
 
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You have an odd number of "x"s outside the square root so try this: Write the integral as \int x^2\sqrt{36- x^2}(x dx) and let u= 36- x2. Then du= 2x dx so x dx= (1/2)du and x^2= 36- u.
 

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