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

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In summary: Dthanks all for your help!In summary, the conversation discusses the process of finding the integral of a given equation with limits of 1 and 0. The individual seeks confirmation on their method for solving the first part of the equation and is advised to use a trigonometric substitution technique. After correcting an error, the individual successfully solves the equation using the suggested substitution and thanks the others for their help.
  • #1
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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  • #2
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.
 
  • #3
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:
  • #4
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
 
  • #5
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:
 
  • #6
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
 

1. What is the integral of x(sqrt(1-x^2)) + 1/2 -x^2/2?

The integral of x(sqrt(1-x^2)) + 1/2 -x^2/2 is equal to (-1/3)(1-x^2)^3/2 + x^2/2 + x + C, where C is the constant of integration.

2. How do you solve this integral?

To solve this integral, you can use the substitution method by letting u = 1-x^2. Then, du = -2x dx and the integral becomes -1/2 * integral of u^(1/2) + 1/2 du. Using the power rule, the integral becomes -1/2 * (2/3)u^(3/2) + u/2 + C. Substituting back in for u and simplifying, we get the final answer of (-1/3)(1-x^2)^3/2 + x^2/2 + x + C.

3. Is this the correct answer?

Yes, the answer (-1/3)(1-x^2)^3/2 + x^2/2 + x + C is the correct answer to the integral of x(sqrt(1-x^2)) + 1/2 -x^2/2. However, keep in mind that there may be variations in the answer depending on the method used and the constant of integration.

4. Can you explain the steps to solving this integral?

Sure. To solve this integral, we first use the substitution method by letting u = 1-x^2. Then, we find the value of du in terms of x and substitute it into the integral. This allows us to simplify the integral to a form that we can easily integrate using the power rule. Once we have integrated the simplified form, we substitute back in for u and simplify to get the final answer.

5. Are there any other methods to solve this integral?

Yes, there are other methods to solve this integral such as integration by parts or using trigonometric substitutions. However, the substitution method is often the easiest and most straightforward approach for this particular integral.

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