Integral of x(sqrt(1-x^2)) + 1/2 -x^2/2 IS IT RIGHT?

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Homework Help Overview

The discussion revolves around the integral of the expression x(sqrt(1-x^2)) + 1/2 - x^2/2, evaluated from 0 to 1. Participants are examining the correctness of the original poster's approach and calculations, particularly focusing on the first term of the integral.

Discussion Character

  • Exploratory, Assumption checking, Mathematical reasoning

Approaches and Questions Raised

  • The original poster attempts a trigonometric substitution method to solve the integral but questions the validity of their calculations. Some participants suggest checking the differentiation of the result to confirm correctness. Others point out an early error in the substitution process and propose alternative methods that may simplify the problem.

Discussion Status

Participants are actively engaging with the original poster's calculations, providing feedback on errors and suggesting alternative approaches. There is a recognition of the original poster's confusion regarding the limits of integration, and some participants express that they have arrived at different results, indicating a variety of interpretations being explored.

Contextual Notes

There is mention of a potential error in the original poster's substitution, as well as a suggestion to use a different substitution method that may yield a quicker solution. The discussion reflects a learning environment where participants are encouraged to explore and clarify their understanding of the integral.

indie452
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Homework Statement



integral of [ x*sqrt[1-x2] + 1/2
- x2/2] dx - with the limits of 1 and 0

the last two bits are easy but i just wanted to confirm what i have done for the first bit is correct...ive tried using a method that takes less time but i don't know if its right...

Homework Equations



sin(x)^2 + cos(x)^2 = 1

cos(arcsinx) = sqrt(1-x2)

The Attempt at a Solution



for the first bit:

x = sinu
dx = cosu du
1-x2 = cos(u)^2

so it becomes: sinu(cos(u)2)cosu du

sinu du = -dcosu

so :- -cos(u)3 dcosu
integrate this beacomes: -cos(u)4/4

which then becomes
-cos(arcsinx)4/4 which is just: -(1-x2)2/4

so putting this together with the 2nd and 3rd part gives

-(1-x2)2/4 + x/2 - x3/6 with limts 1&0

which equals 1/3

so is this right?
 
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Use the right tool for the right situation. Trigonometric substitution certainly works, but it takes you on quite a detour. To check if your answer is correct you simply differentiate it so let's do that.

[tex]\frac{d}{dx} \frac{-(1-x^2)^2}{4}=x(1-x^2) \neq x \sqrt{1-x^2}[/tex]

You make an error very early on after you state that 1-x^2=cos(u)^2 you conclude in the next line that [itex]\sqrt{1-x^2}=\cos^2u[/itex]. In other words you forgot to take the square root. The rest seems to be correct.

There is however a method that is way easier take u=1-x^2 and you're done.
 
as soon as you pointed out that error early on i felt like kicking myself :rolleyes: yeah i see what to do now

(do you still get 1/3 as the answer cause I've just redone it and I am still getting the same answer with the limits in)

thanks
 
Last edited:
For the first term I get 1/3 for the second 1/2 and for the third -1/6. So 1/3+1/2-1/6=2/3
 
Welcome to PF!

Hi indie452! Welcome to PF! :smile:

(have a square-root: √ :wink:)
Cyosis said:
There is however a method that is way easier take u=1-x^2 and you're done.

indie452, have you tried Cyosis's :smile: sub, u = √(1 - x2)?

It's much quicker, and it's a standard technique that you do need to learn. :smile:
 
yeah i tried it ... the reason i got the wrong answer the 2nd time was cause i messed up on subbing the limits but i got the right answer now
 

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