Can You Simplify Integrals by Separating Radicals?

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SUMMARY

The discussion centers on the incorrect application of integral separation involving radicals. Specifically, the user inquires about breaking up integrals such as ∫(1)/(sqrt(16-9x²)³) dx and ∫(x²)/(sqrt(x²-9)) dx. The consensus is that this approach is invalid due to the property that √(a - b) does not equal √a · √(-b). Participants emphasize the importance of understanding radical properties before attempting such separations.

PREREQUISITES
  • Understanding of integral calculus
  • Familiarity with properties of radicals
  • Knowledge of integral separation techniques
  • Basic algebraic manipulation skills
NEXT STEPS
  • Review properties of radicals in calculus
  • Study integral separation techniques in detail
  • Practice solving integrals involving square roots
  • Explore advanced integration methods such as substitution and integration by parts
USEFUL FOR

Students studying calculus, particularly those focusing on integral calculus and anyone seeking to clarify the rules governing the manipulation of integrals involving radicals.

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Homework Statement


for the following integrals, am I allowed to break them up like so:

1. ∫(1)/(sqrt(16-9x²)³) dx

= ∫(1)/(√16)³ · ∫(1)/(√-9x²)³ dx

2. ∫(x²)/(sqrt(x²-9)) dx

= ∫(x²)/(√x²) · ∫(x²)/(√-9) dx

3. ∫(1)/(x²(sqrt(a²+x²))) dx

= ∫(1)/(x²) · ∫(1)/(√a²) · ∫(1)/(√x²) dx

? ? ?


Homework Equations


none


The Attempt at a Solution


I need to know if I'm allowed to break them up like this before I start attempting a solution
 
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Is this what you are writing for #1?
\int \frac{1}{\left( \sqrt{16-9x^2}\right)^3} dx
=\int \frac{1}{\left( \sqrt{16}\right)^3} dx \cdot \int \frac{1}{\left( \sqrt{-9x^2}\right)^3} dx
Yikes. No, you cannot do that!

\sqrt{a - b} \ne \sqrt{a} \cdot \sqrt{-b}
Better review the properties of radicals.
 
Ok, guess i'll try something else. Thanks
 

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