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HSC 2015 MX2 Integration Marathon (archive) (1 Viewer)
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Re: MX2 2015 Integration Marathon
So it's related to an existing question. I couldn't tell.
Paradoxica
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Re: MX2 2015 Integration Marathon
If you people needed a hint, break up the numerator (Using your intuition, not just willy-nilly!) and make a substitution, or manipulate the form.a) Differentiate
b) Hence or otherwise, solve
\sqrt{1-x^2}}e^x}dx)
Computer Algebra Systems don't always find the elementary expression of an integrand, as there are integrals I know of which Wolfram Alpha cannot solve, but humans can. This is my go to place for these integrals. (P.S. I have quite a few left which I will continue to post as we progress)
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leehuan
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Paradoxica
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Re: MX2 2015 Integration Marathon
Are you certain a minor mistake doesn't exist? By differentiating my answer and also looking at my notes, the integral can certainly be evaluated.
Ekman
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Re: MX2 2015 Integration Marathon
e^x}{(1-x)\sqrt{1-x^2}} dx = \int \frac{(1-x^2)e^x}{(1-x)\sqrt{1-x^2}} dx + \int \frac{e^x}{(1-x)\sqrt{1-x^2}} dx )
}{(1-x)\sqrt{1-x^2}} \right) = \frac{1}{(1-x)\sqrt{1-x^2}} )
e^x}{(1-x)\sqrt{1-x^2}} dx = \frac{(1-x^2)e^x}{(1-x)\sqrt{1-x^2}} + c )
 = \frac{(1-x^2)}{(1-x)\sqrt{1-x^2}} )
Im not sure where the substitution comes from but this is what I did:
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Drsoccerball
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Re: MX2 2015 Integration Marathon
Heres our over 70 for BOS trials.Im not sure where the substitution comes from but this is what I did:
Paradoxica
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sirable1
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Re: MX2 2015 Integration Marathon
Fairly easy goer here.
Solve:
Fairly easy goer here.
Solve:
Ekman
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Re: MX2 2015 Integration Marathon
 )
} +c = \tan^{-1}{\left(x+\frac{1}{x}\right)} + c )
This is similar to a previous question you asked.Ahah yeah probs :L
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Drsoccerball
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Re: MX2 2015 Integration Marathon
u=(x+3) and then you have an ln and an inverse tan.
LOL this is more my levelFairly easy goer here.
Solve:
u=(x+3) and then you have an ln and an inverse tan.
Paradoxica
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Re: MX2 2015 Integration Marathon
Manipulate both until you can make the substitution
Then for both integrals, make the appropriate trigonometric substitutions. For the second one, it can be split so the inverse secant is acquired immediately, and the other part of the integral is handled by trigonometry.
As a bonus fact, most Computer Algebra Systems fail to evaluate most of the known "hard" integrals that are handled by symmetric polynomial substitutions.
Ahahahaha.... We both know what to do here it seems.Hey Paradoxica, here are a few integrals for your integration bank:
I have a feeling you will be able to solve them
Manipulate both until you can make the substitution
As a bonus fact, most Computer Algebra Systems fail to evaluate most of the known "hard" integrals that are handled by symmetric polynomial substitutions.
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Paradoxica
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Re: MX2 2015 Integration Marathon
A Combination of Intuition, Experience, Hindsight and Luck. Symbolic Integration is an Art, not a Science. Numerical Integration is a Science. Specifically a Computer Science. Okay I'll stop Misusing Definitions now.How do people know to substitute something like:
Re: MX2 2015 Integration Marathon
I see the consequence of it (Ekman's soln) after some manipulation of the integrand but I've seen people use it straight away. I guess they've completed the manipulation mentally..
porcupinetree
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Re: MX2 2015 Integration Marathon
Although I do quite enjoy reading Ekman's and Paradoxica's solutions to some of their harder integrals, I admit.
Finally a question which isn't completely nutsFairly easy goer here.
Solve:
Although I do quite enjoy reading Ekman's and Paradoxica's solutions to some of their harder integrals, I admit.
Paradoxica
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Re: MX2 2015 Integration Marathon
With enough practise, you learn to be able to do that in your head. It's like mental arithmetic, only vastly more impressive, because there are plenty of things that can go wrong, and no methodical process for arbitrary symbolic integration.
Re: MX2 2015 Integration Marathon
Yeah, the mathematical intuition required to see that division is quite spectacular!
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