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Hi there guys I don't know if the title of the question should be the one for this but the thing is that I'm trying to solve this integral $\int \frac {\frac 12-u^2}{2u^4-2u^2+1}$$du$ and I have this doubt where if I try to do this $2(u^2)^2-2(u)^2+1$ in the denominator, I can't see how to factor it, plus I've had completed the square from what I've already tried to do a trigonometric substitution but seems that is not the best way to go since it gets way complicated, I see the term "irreducible factorization" from wolfram alpha when I try to factor it from there but that's something I don't know about yet, so it is possible to do this without that "irreducible factorization"? If so, what do I need to know? I do know the ways of integration but when I find problems like this one where you can't factor a polynomial like this one I get stuck really bad, I hope this question isn't too silly or something I'm just really trying to learn this real bad and I don't know somewhere else to ask!

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  • $\begingroup$ The denominator factorises to $2[(u^{2} - 1/2)^{2} + 1/4]$. That might imply there is a substitution you can make. $\endgroup$ – Mattos Feb 10 '16 at 3:11
  • $\begingroup$ double the denominator. What is another way to write that? $\endgroup$ – Will Jagy Feb 10 '16 at 3:11
  • $\begingroup$ @Mattos thanks for your answer and that's what I also got when doing the square completion but all I can think of is going to trigonometric substitution from there and it gets worse after doing that or is there other substitution that i'm not noticing? $\endgroup$ – CryoCodex Feb 10 '16 at 3:49
  • $\begingroup$ Although I agree that it's not trivial to factor the denominator, it does in fact factor (as all polynomials of degree higher than $2$ must), as $2(x^2+x\sqrt{1+\sqrt2}+1/\sqrt2)(x^2-x\sqrt{1+\sqrt2}+1/\sqrt2)$. $\endgroup$ – Greg Martin Feb 10 '16 at 7:24
  • $\begingroup$ @GregMartin Thanks man! but how did you factor it in such way if I may ask? because i'm not seeing it how it should factor with the conventional ways I know this may be a silly question so, i'm sorry if it is $\endgroup$ – CryoCodex Feb 10 '16 at 13:17
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I hope that you will be given a simpler way. Meanwhile, I give you my approach.

For the time being, let me set $u^2=x$ which makes the integrand $$\frac{1-2 x}{2 \left(2 x^2-2 x+1\right)}$$ Solving $2x^2-2x+1=0$ makes $$2(2x^2-2x+1)=4(x-a)(x-b)$$ with $a=\frac{1-i}2$ and $b=\frac{1+i}2$. Now, partial fraction decomposition makes $$\frac{1-2 x}{2 \left(2 x^2-2 x+1\right)}=\frac{1-2 x}{4(x-a)(x-b)}=\frac{1-2 a}{4 (a-b) (x-a)}+\frac{2 b-1}{4 (a-b) (x-b)}$$ that is to say $$\frac{1-2 x}{2 \left(2 x^2-2 x+1\right)}=\frac A {x-a}+\frac B {x-b}$$ where, after simplifications $A=B=-\frac 14$.

Now, back to $u$, we then have $$ \frac {\frac 12-u^2}{2u^4-2u^2+1}=-\frac 14\Big(\frac 1 {u^2-a}+\frac 1 {u^2-b}\Big)$$ and partial fraction decomposition can again be done.

So, by the end, you will just need to integrate four terms $\frac 1 {u+\alpha_i}$ where the $\alpha_i$'s are complex numbers.

I am sure that you can take it from here.

Edit

If you give the integral to Wolfram Alpha, it will give $$I=\int\frac {\frac 12-u^2}{2u^4-2u^2+1}\,du=-\frac{\tan ^{-1}\left(\frac{u}{\sqrt{-\frac{1}{2}-\frac{i}{2}}}\right)}{2 \sqrt{-2-2 i}}-\frac{\tan ^{-1}\left(\frac{u}{\sqrt{-\frac{1}{2}+\frac{i}{2}}}\right)}{2 \sqrt{-2+2 i}}$$ whci is just a linear combination of logarithms with complex terms since it rewrite $$I=\frac{1}{4} \sqrt{1+i} \left(\tanh ^{-1}\left(u\sqrt{1+i} \right)-(-1)^{3/4} \tanh ^{-1}\left(u\sqrt{1-i} \right)\right)$$ with $$\sqrt{1\pm i}=\sqrt[4]{2} \Big(\cos \left(\frac{\pi }{8}\right)\pm i \sin \left(\frac{\pi }{8}\right)\Big)$$

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  • $\begingroup$ If you set $u^2=x$, then you need to change $dx$ to $2u\,du$. $\endgroup$ – Greg Martin Feb 10 '16 at 7:19
  • $\begingroup$ I only used $u^2=x$ just for the simplification of the integrand and to show that we just face a sum of integrals looking like $\int\frac {du}{u+\alpha_i}$ where the $\alpha_i$'s are complex numbers. I did not use $du$ or $dx$ anywhere. But, for sure, I may be wrong ! $\endgroup$ – Claude Leibovici Feb 10 '16 at 7:27
  • $\begingroup$ Fair enough, I guess. $\endgroup$ – Greg Martin Feb 10 '16 at 7:28
  • $\begingroup$ @GregMartin. Basically, I did the same as in your comment (I think). Cheers. $\endgroup$ – Claude Leibovici Feb 10 '16 at 7:30
  • $\begingroup$ Thanks man! I really appreciate your help but as far I'm aware of in my country you don't get to see complex numbers until calculus IV so I just don't get what you did at all also, shouldn't be $ 2x^2-2x+1=0 $ instead of $ 2x^2-2x-1=0 $ ? $\endgroup$ – CryoCodex Feb 10 '16 at 13:10
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In a comment, you wondered if the problem could be instead $$\int \frac {\frac 12-u^2}{2u^4-2u^2-1}\, du$$ Doing the same as before, we have $$2x^2-2x-1=2(x-a)(x-b)$$ with $$ \quad a=\frac{1}{2} \left(1-\sqrt{3}\right)<0\quad,\quad b=\frac{1}{2} \left(1+\sqrt{3}\right)>0$$ and, again, partial fraction decomposition makes$$\frac{1-2 x}{2 \left(2 x^2-2 x-1\right)}=\frac A {x-a}+\frac B {x-b}$$ with $A=\frac 14$, $B=-\frac 14$.

Back to $u$, $$\frac {\frac 12-u^2}{2u^4-2u^2-1}=\frac 14\Big(\frac 1 {u^2-a}-\frac 1 {u^2-b}\Big)$$ Now, we face simple integrals and the final result is simply $$I=\frac{1}{4} \left(\sqrt{\sqrt{3}-1} \tanh ^{-1}\left(u\sqrt{\sqrt{3}-1} u\right)-\sqrt{1+\sqrt{3}} \tan ^{-1}\left(u\sqrt{1+\sqrt{3}} \right)\right)$$

Taking into account your remark in a comment, there is probably one more typo in the textbook.

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