- #36
mitochan
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Please take a look at attached calculation.bob012345 said:Is there some simple easy trick I am missing that makes this much easier than ? You can tell me privately if you wish.
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Please take a look at attached calculation.bob012345 said:Is there some simple easy trick I am missing that makes this much easier than ? You can tell me privately if you wish.
Thanks. It does seem simpler with the ##1## in the denominator.mitochan said:Please take a look at attached calculation.
But for x=3i makes some of the denominators of the identity zero...ergospherical said:@bob012345, multiply the identity ##\dfrac{1}{(x+3i)(x-3i)} = \dfrac{A}{x + 3i} + \dfrac{B}{x-3i}## by the factor ##(x-3i)##, and then set ##x = 3i## (you can do this because the identity holds for all ##x##). And do a similar thing to find ##A##.
If you multiply both sides by one of the factors, say ##x-3i##, solve for ##A## and then set ##x=3i## it works. Cancel those terms before you make the substitution.Delta2 said:But for x=3i makes some of the denominators of the identity zero...
That's the idea, so you can find A, then do x=-3i and find B.Delta2 said:But for x=3i makes some of the denominators of the identity zero...
I think it is being suggested this is like the proof ##1=2##.Tapias5000 said:That's the idea, so you can find A, then do x=-3i and find B.
Emmm, by multiplying both sides by (x-3i) and then making x=3i, you will not fall into any error, it is valid to do so.bob012345 said:If you multiply both sides by one of the factors, say ##x-3i##, solve for ##A## and then set ##x=3i## it works. Cancel those terms before you make the substitution.
You can think of it like this:Delta2 said:But for x=3i makes some of the denominators of the identity zero...
This is exactly how I would do it.mitochan said:Please take a look at attached calculation.
No, not at all. I think you might be referring to the post by ergospherical.bob012345 said:I think it is being suggested this is like the proof ##1=2##.
An important point is that the equation above is an identity: one that is is true for all values of the variable x, excepting only ##x = \pm 3i##. For any values of x other than these two, we can multiply both sides of the equation by ##(x + 3i)(x - 3i)## to get this equation:ergospherical said:multiply the identity ##\dfrac{1}{(x+3i)(x-3i)} = \dfrac{A}{x + 3i} + \dfrac{B}{x-3i}## by the factor ##(x-3i)##, and then set ##x = 3i## (you can do this because the identity holds for all ##x##). And do a similar thing to find ##A##.
hehe, I use Quora more so my answers are usually complete aum, I'm not used to this forum at all hehe.Delta2 said:It should not have been left to the reader to do first the multiplication of the identity by x-3i, it should have been explicitly mentioned by @ergospherical :D.
Thenbob012345 said:Thanks. It does seem simpler with the ##1## in the denominator.
I just used a basic trig with ##tan(\theta)= \large\frac{x}{a}## and it worked fine.mitochan said:Then
[tex]\int \frac{dx}{x^2+2}= \frac{1}{\sqrt{2}}\int\frac{d\frac{x}{\sqrt{2}}}{(\frac{x}{\sqrt{2}})^2+1}=...[/tex]
The image in post #36 shows why this substitution works. For ##x^2 + 2##, the length of the leg adjacent to the angle is ##\sqrt 2## rather than 1.bob012345 said:I just used a basic trig with ##tan(\theta)= \large\frac{x}{a}## and it worked fine.
Of course and the point I made was the method I used works for any ##a##. That's why I used ##a## and not ##1##. I found it easier than manipulating the integral first to make it a ##1## or using complex numbers.Mark44 said:The image in post #36 shows why this substitution works. For ##x^2 + 2##, the length of the leg adjacent to the angle is ##\sqrt 2## rather than 1.
I tried it and it was not easier by any means.vela said:Now try it with a hyperbolic trig substitution.