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Integration problems: Intriguing Integrals

  • Thread starter BioCore
  • Start date
  • #1
BioCore
1. Homework Statement
[tex]\int[/tex]x^3[tex]\sqrt{}a^2-x^2[/tex]dx


2. Homework Equations



3. The Attempt at a Solution
I came this far with my solution:

a^5[tex]\sqrt{}sin^3\Theta cos^2\Theta[/tex]d[tex]\Theta[/tex]
a^5[tex]\sqrt{}sin^3\Theta-sin^5\Theta[/tex]d[tex]\Theta[/tex]

The answer given to me is:

-1/15(a^2 - x^2)^3/2 (2a^2 + 3x^2) + C

It seems that they used sine theta, and not cos theta. I have currently tried to do that but am not sure how I would get rid of the extra sin[tex]\Theta[/tex].

Thanks for the help.
1. Homework Statement



2. Homework Equations



3. The Attempt at a Solution
 
Last edited by a moderator:

Answers and Replies

  • #2
rock.freak667
Homework Helper
6,230
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[tex]\int x^3 \sqrt{a^2-x^2}dx[/tex]

let [itex]x=acos\theta \Rightarrow dx=-asin\theta[/itex]

[tex]\int x^3 \sqrt{a^2-x^2}dx \equiv \int (acos\theta)^3\sqrt{a^2-a^2cos^2\theta} \times -asin\theta d\theta[/tex]

[tex]\int - a^4cos^3\theta \sqrt{a^2(1-cos^2\theta)} \times -asin\theta d\theta[/tex]

[tex]-a^5 \int cos^3\theta sin^2\theta d\theta[/tex]

recall that [itex]sin^2\theta +cos^2\theta=1[/itex]

and substitute for [itex]sin^2\theta[/itex]
 
  • #3
BioCore
[tex]\int x^3 \sqrt{a^2-x^2}dx[/tex]

let [itex]x=acos\theta \Rightarrow dx=-asin\theta[/itex]

[tex]\int x^3 \sqrt{a^2-x^2}dx \equiv \int (acos\theta)^3\sqrt{a^2-a^2cos^2\theta} \times -asin\theta d\theta[/tex]

[tex]\int - a^4cos^3\theta \sqrt{a^2(1-cos^2\theta)} \times -asin\theta d\theta[/tex]

[tex]-a^5 \int cos^3\theta sin^2\theta d\theta[/tex]

recall that [itex]sin^2\theta +cos^2\theta=1[/itex]

and substitute for [itex]sin^2\theta[/itex]
I would assume that using your solution I would have the right answer. The only problem is that in the solutions my Professor gave us, he has an intermediate step before the final solution such as this:

[tex]a^5 \int sin^3\theta cos^2\theta d\theta[/tex]

Also in explaining to us Trig Substitution he placed [tex]\sqrt{a^2 - x^2}[/tex] as the bottom portion of the right triangle. Thus he stated that

x = asin[tex]\Theta[/tex]
dx = acos[tex]\Theta[/tex]
[tex]\sqrt{a^2 - x^2}[/tex] = acos\theta
 
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  • #4
rock.freak667
Homework Helper
6,230
31
I would assume that using your solution I would have the right answer. The only problem is that in the solutions my Professor gave us, he has an intermediate step before the final solution such as this:

[tex]a^5 \int sin^3\theta cos^2\theta d\theta[/tex]

Also in explaining to us Trig Substitution he placed [tex]\sqrt{a^2 - x^2}[/tex] as the bottom portion of the right triangle. Thus he stated that

x = asin[tex]\Theta[/tex]
dx = acos[tex]\Theta[/tex]
[tex]\sqrt{a^2 - x^2}[/tex] = acos\theta
Your teacher used x=sin[itex]\theta[/itex], I used x=cos[itex]\theta[/itex].
So in his triangle, [itex]sin\theta=\frac{x}{a}[/itex] which would make the adjacent side,[itex]\sqrt{a^2-x^2}[/itex]. At the end of that integral you would get terms involving [itex]cos\theta[/itex]

In my substitution, the opposite side would be [itex]\sqrt{a^2-x^2}[/itex] and at the end of my integral I would have terms involving [itex]sin\theta[/itex]

So in theory it should all work out the same.
 
  • #5
BioCore
Ok following your steps:

[tex]-a^5 \int \sqrt{1 - sin^2\theta} (1 - sin^2\theta) (sin^2\theta)d\theta[/tex]

= [tex]-a^5 \int (1 - sin^2\theta)^3/2 (sin^2\theta)d\theta[/tex]

= [tex]-a^5 \int sin^3\theta - sin^6\theta d\theta[/tex]

= [tex]-a^5 [1/4sin^4\theta + 1/7sin^7\theta ][/tex] +C

Now I know that I should substitute for the sine function, but is this so far correct?
 
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