(adsbygoogle = window.adsbygoogle || []).push({}); 1. The problem statement, all variables and given/known data

[tex] \int^{1}_{-1} 2x\sqrt{(1-x^{2})^{3} +1} dx [/tex]

3. The attempt at a solution

This is one of those questions in a math book that is almost never assigned and can be seen as "going above and beyond" what the chapter teaches you. I kind of improvised on the solution by substituting variables twice throughout the problem, so I am not sure if the solution is correct. I know that this is a definite integral but I only solved the indefinite portion of it, at least I think that I did. If I can get someone to give me the thumbs up on the indefinite solution I'll plug in the numbers for fun, but this is the part that counts right? ha, anyway:

Let:

[tex] u = 1-x^{2} [/tex]

[tex] du = -2x dx [/tex]

[tex] dx = -\frac{1}{2}x^{-1} du [/tex]

So now the problem is going to look like so:

[tex] \int^{1}_{-1} -\frac{1}{2} \frac{2x}{x} \sqrt{(u)^{3}+1}du [/tex]

and then:

[tex] \frac {2x}{x} = 2 \rightarrow -\frac{1}{2}(2) = -1 [/tex]

so now its:

[tex] \int^{1}_{-1} - \sqrt{(u)^{3}+1} du [/tex]

or better yet:

[tex] \int^{1}_{-1} -((u)^{3}+1)^{1/2} du [/tex]

and now it gets to the part of where I am not sure if this is legal to do:

Let:

[tex] s = (u)^{3}+1 [/tex]

[tex] ds =3u^{2} du[/tex]

[tex] du = \frac{1}{3}u^{-2} ds [/tex]

and since:

[tex] s = (u)^{3} +1 \rightarrow u = (s-1)^{1/3} = (s^{1/3}-1)[/tex]

so now the problem looks like:

[tex] \int^{1}_{-1} -\frac{1}{3}(s)^{1/3}(u)^{-2} ds [/tex]

which is the same as:

[tex] \int^{1}_{-1} -\frac{1}{3}(s)^{1/3}(s^{1/3}-1)^{-2} ds [/tex]

and then to get rid of the -2 exponent:

[tex] \int^{1}_{-1} -\frac{1}{3}(s)^{1/3}(s^{-2/3}-1) ds [/tex]

then distribute out the binomial:

[tex] \int^{1}_{-1} -\frac{1}{3}(s^{-1/3} - s^{1/2}) ds [/tex]

finally distribute the -1/3:

[tex] \int^{1}_{-1} -\frac{1}{3}s^{-1/3} + \frac{1}{3}s^{1/2} ds [/tex]

now lets take the anti-derivative of what we have:

[tex] -\frac{1}{2}(s)^{2/3}+\frac{2}{9}(s)^{3/2} [/tex]

and then substitute:

[tex] -\frac{1}{2}(u^{3}+1)^{2/3}+\frac{2}{9}(u^{3}+1)^{3/2} [/tex]

and once more:

[tex] -\frac{1}{2}((1-x^{2})^{3}+1)^{2/3}+\frac{2}{9}((1-x^{2})^{3}+1)^{3/2} [/tex]

then finally to complete the indefinite integral of the problem:

[tex] -\frac{1}{2}((1-x^{2})^{3}+1)^{2/3}+\frac{2}{9}((1-x^{2})^{3}+1)^{3/2} + C[/tex]

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# Homework Help: Extra Credit Definite Integral - No Solution In Book- Did I do This right?

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