Calculus II - Trigonometric Substitutions - Evaluate integral dx/sqrt(1-2x^2)

1. Aug 3, 2011

GreenPrint

1. The problem statement, all variables and given/known data

Evaluate
integral dx/sqrt(1-2x^2)

2. Relevant equations

3. The attempt at a solution

See my work in attachment
http://www.wolframalpha.com/input/?i=integral+dx%2Fsqrt%281-2x^2%29
Code (Text):
>> syms x
>> int(1/sqrt(1-2*x^2))

ans =

(2^(1/2)*asin(2^(1/2)*x))/2
I know how I chose to evaluate this integral is not the traditional way of how things are done but I strongly believe that what I did is mathematically correct as you can see in my steps and so I should get the same answer but I don't, I don't know what I'm doing wrong and hope somebody can tell me...

Attached Files:

• Scan.jpg
File size:
15.6 KB
Views:
245
Last edited: Aug 3, 2011
2. Aug 3, 2011

vela

Staff Emeritus
I didn't check your work closely, but I didn't see anything wrong.

Remember that the antiderivatives can differ by a constant.

3. Aug 3, 2011

GreenPrint

-cos^(-1)(sqrt(2)x)/sqrt(2) + c
is equal to the what woflram alpha says to be the answer
(sin^(-1)(sqrt(2) x))/sqrt(2)+constant

I hate how they can differ by a constant and that i can get the right answer but have it appear as the wrong one because of that darn constant and was wondering how do i check that the my answer is indeed correct?

4. Aug 3, 2011

vela

Staff Emeritus
The easy way to check your answer is to differentiate it and see if you get the integrand.

5. Aug 3, 2011

GreenPrint

It would appear as if it is correct ^_^ I still don't like this though >_> so there are multiple antiderrivitives that are correct because of the constant @_@ so i guess like if i answered this question on the test a professor couldn't mark it wrong because it's actually right even though it doesn't match with (sin^(-1)(sqrt(2) x))/sqrt(2)+constant, there would have to be multiple answers on the answer key i guess?

6. Aug 3, 2011

Bohrok

You can look at the different answers coming from the constant, or from another places. If I solve ∫1/√(1-x2) dx with the substitution x = sinθ, I'll get sin-1x + c. If I use x = cosθ, I'll get cos-1x + c.* And because of the identity $\sin^{-1}x + \cos^{-1}x = \pi/2 \Rightarrow \sin^{-1}x = -\cos^{-1}x + \pi/2$, your answer and the other answers are the same once that $\pi/2$ is absorbed into the constant. Also, if you differentiate both sides of the second equation, both sides will be the same. You can do that to show the two "different" answers to your problem are the same.

*Same kind of thing happens if you use trig substitutions other than the ones normally used, and they'll give you basically the same answer, except maybe for using cot x instead of tan x because of the part of its domain used for having an inverse. At least in the case with arcsin and arccos, they are two equivalent answers when they turn up as antiderivatives.

Last edited: Aug 3, 2011
7. Aug 3, 2011

tiny-tim

Hi GreenPrint!

cos-1(A) = π/2 - sin-1(A)