Integration of 1/x^2 sqrt(16-x) with substitution

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Homework Help Overview

The discussion revolves around the integration of the function \( \frac{1}{x^2 \sqrt{16 - x^2}} \) using a substitution method. The original poster attempts to apply the substitution \( x = 4 \sin y \) to simplify the integral.

Discussion Character

  • Exploratory, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the substitution method and the resulting integral, with the original poster expressing uncertainty about how to proceed after transforming the integral into a form involving \( \frac{1}{\sin^2 y} \). Others mention the relationship between \( \frac{1}{\sin^2 y} \) and \( \csc^2 y \) and reference the integral of \( \csc^2 y \).

Discussion Status

The discussion is ongoing, with participants providing hints and references to known integrals. There is no explicit consensus on the next steps, but some guidance has been offered regarding the integral of \( \csc^2 y \).

Contextual Notes

The original poster indicates a lack of confidence in integrating \( \frac{1}{\sin^2 y} \), which may suggest a gap in their understanding of integral calculus concepts related to trigonometric functions.

teng125
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my question is integra 1 / x^2 sqr root(16-x^2)
i use x = 4 sin y
then i got to integrate 1/16 times [1/(sin y)^2] in which i got stuck.pls help...how do i continue from there as i not sure ow to integra 1/(siny^2
 
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[itex]\frac{1}{sin^2y}= csc^2y[/itex]- you should know the integral for that. What is the derivative of cot y?

(Sorry- I had left out a "}")
 
Last edited by a moderator:
Since I can't see Halls' Latex image, he was saying that [tex]\int \csc^2(y)dy[/tex] is very simple, noting that what the derivative of cot(y) is.
 
Ivy's image is: [tex]\frac{1}{sin^2y}= csc^2y[/tex]
 

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