Calculus with inverse trig functions

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Homework Help Overview

The discussion revolves around evaluating the integral of the function (1/Sqrt(5x-x^2)), which involves inverse trigonometric functions and integration techniques in calculus.

Discussion Character

  • Exploratory, Assumption checking, Mathematical reasoning

Approaches and Questions Raised

  • Participants explore the method of completing the square for the expression 5x-x^2, with some expressing confusion regarding the constant 5 in the equation.

Discussion Status

Some participants have attempted to manipulate the expression into a more workable form, while others have acknowledged the challenges posed by the specific coefficients involved. There is an ongoing exploration of how to find the necessary constants for completing the square.

Contextual Notes

There is mention of a specific form required for completing the square, and participants are questioning how to derive the constant 'k' in the transformation of the expression.

famallama
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Homework Statement


Evaluate the integral of (1/Sqrt(5x-x^2))


Homework Equations


[d/dx]{arcsin(x)}=(du/dx)/sqrt(1-x^2)


The Attempt at a Solution


arcsine(2x-5)/5


I did end up getting the right answer, but have no idea how I got there.
 
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So we have

[tex]\int \frac{1}{\sqrt{5x-x^2}}[/tex]

try completing the square for 5x-x2 (i.e. put it into the form a[x+h]2+k)
 
I have tried that, but the 5 is what is throwing me off.
 
famallama said:
I have tried that, but the 5 is what is throwing me off.

5x-x2 = -(x2-5x) = -(x+5/2)2+k. Can you find 'k'?
 
thank you very much
 

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