How to Solve the Integral of 1/sqrt(7-x^2) – A Step-by-Step Guide

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Discussion Overview

The discussion revolves around solving the integral of \( \frac{1}{\sqrt{7-x^2}} \). Participants explore various steps in the integration process, including substitution methods and the application of trigonometric identities. The focus is on the mathematical reasoning and techniques involved in evaluating the integral.

Discussion Character

  • Mathematical reasoning
  • Homework-related

Main Points Raised

  • One participant presents the integral \( I=\int \frac{dx}{\sqrt{7-x^2}} \) and claims the solution involves arcsine, specifically \( \arcsin\left(\frac{\sqrt{7}x}{7}\right)+C \).
  • Another participant describes their substitution \( x=\sqrt{7}\sin(u) \) and the corresponding differential \( dx=\sqrt{7}\cos(u) \, du \), leading to a transformation of the integral.
  • There is a discussion about the correctness of the steps taken, with one participant expressing uncertainty about their solution but later confirming it as correct.
  • Participants engage in clarifying the process, with one suggesting that earlier communication of the solution would have been beneficial.

Areas of Agreement / Disagreement

Participants do not reach a consensus on the clarity of the solution process, with some expressing uncertainty about the correctness of their steps while others affirm the validity of the approach. The discussion remains somewhat unresolved regarding the best method to present the solution.

Contextual Notes

Some participants express confusion over the steps involved in the integration process, indicating potential limitations in their understanding of the substitution method and its implications for the integral's evaluation.

karush
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242.ws3.2
$\displaystyle
I=\int \frac{dx}{\sqrt{7-x^2}}
=\arcsin\left(\frac{\sqrt{7}x}{7}\right)+C$
I know this is a standard Integral but was given to solve it
$x=\sqrt{7}\sin\left({u}\right) \therefore dx=\sqrt{7}\cos\left({u}\right) \, du$
I proceeded but 😰
 
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karush said:
242.ws3.2
$\displaystyle
I=\int \frac{dx}{\sqrt{7-x^2}}
=\arcsin\left(\frac{\sqrt{7}x}{7}\right)+C$
I know this is a standard Integral but was given to solve it
$x=\sqrt{7}\sin\left({u}\right) \therefore dx=\sqrt{7}\cos\left({u}\right) \, du$
I proceeded but 😰

putting $x=\sqrt{7}\sin\left({u}\right) \therefore dx=\sqrt{7}\cos\left({u}\right) \, du$
we have
$\sqrt{7-x^2} = \sqrt{7}\cos (u)$
so integal becomes
$I=\int \frac{dx}{\sqrt{7-x^2}}= \int \frac{\sqrt{7}\cos\left({u}\right)}{\sqrt{7}\cos\left({u}\right)} \, du= \int du = u +C = \arcsin \frac{x}{\sqrt{7}} + C $
 
really, that's what I did but didn't think it was right... :cool:
 
Then you should have told us that you had a solution and shown how you got it in your first post.
 
$I=\int \frac{dx}{\sqrt{7-x^2}}= \int \frac{\sqrt{7}\cos\left({u}\right)}{\sqrt{7}\cos\left({u}\right)} \, du$
When I got to here I thought it was wrong.
But it was correct😎
 

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