Need help with Integration Problem

[schwampert]

Hi all. So I'm having an issue with the following problem:

We're only doing a review of Calc 1 at this point, it's the beginning of the semester. That being said, I am completely stuck on this problem. I can't figure out how to make integration by parts work, and trying to do integration by substitution just gets me an endless loop where the problem keeps getting more and more complicated. I feel like the answer has something to do with the fact that (1 - 4x⁶)^1/2 could be represented as either 1 - 2x³ or 1 + 2x³, but I'm not sure how to work with that information. Any help would be appreciated.

 

[MarkFL]

Hello, and welcome to FMH!

Consider the substitution:

[MATH]u=2x^3\implies du=6x^2\,dx[/MATH]
And we have:

[MATH]I=\int\frac{1}{\sqrt{1-u^2}}\,du[/MATH]

 

[schwampert]

Thanks for the help.

The answer was staring me right in the face. I was making this way more difficult than it needed to be.

 

[Deleted member 4993]

Thanks for the help.

The answer was staring me right in the face. I was making this way more difficult than it needed to be.

What was your answer?

I = ?

 

[schwampert]

What was your answer?

I = ?

I = sin^-1(2x³)

 

[Deleted member 4993]

I = sin^-1(2x³)

Small (but important correction):

I = sin^-1(2x³) + Constant

 

[Steven G]

(1 - 4x⁶)^1/2 could be represented as either 1 - 2x³ or 1 + 2x³

It seems that you think that \(\displaystyle \sqrt{a-b}\)=\(\displaystyle \sqrt{a}+ \sqrt{b}\) or \(\displaystyle \sqrt{a-b}\)=\(\displaystyle \sqrt{a}- \sqrt{b}\). This is simply not true.
Consider 3 = \(\displaystyle \sqrt{9}=\sqrt{25-16}\neq\sqrt{25}+\sqrt{16} = 5+4 = 9\), not 5 or \(\displaystyle \sqrt{25-16}\neq\sqrt{25}-\sqrt{16} = 5-4 = 1\), not 5