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Re: MX2 Integration Marathon \int \sqrt{1+e^{x}} \ dx

What? I am only giving a hint to him because it seems like he legitimately wants to attempt the question, so I'm helping out. The 2U forums outside of helping other people is dead. In that it isn't like the 4U forums where people share questions for fun, whereas in the 2U forums that doesn't...

Re: HSC 2013 4U Marathon I_n = \int_0^{\frac{\pi}{2}} \cos^{2n} x \ dx $Prove that$ \sum_{k=1}^{n} I_k = (2n+1)I_n - I_0 Its integration but its more about series than anything.

Neither. HINT: $Let the roots of the quadratic equation be$ \ \ \alpha, \beta $The question is asking to prove something to do with k, on the condition that the roots differ by less than 1$ $Translating this into coherent mathematics, means that the quadratic equation has the property...

Re: MX2 Integration Marathon yep if it works, it works. To make that question that I actually differentiated: \frac{1}{2} (\sin^{-1}(e^x) - \cos^{-1}(e^x) + \tan^{-1}(e^x)) So that may be another form for this.

Re: MX2 Integration Marathon I think its better that the Harder 3U questions are asked in the actual 4U marathon imo. Most of the questions there are actually harder 3U if not polynomials

Re: MX2 Integration Marathon The integral resolves into -2\int_0^{\pi/4} \ln(\cos \theta) \ d\theta Which involves Catalan's Constant C C = \sum_{n=0}^{\infty} \frac{(-1)^n}{(2n+1)^2} And because of that I think it will involve Taylor series at some point due to the infinite sum.

Re: MX2 Integration Marathon I don't think its a good integral man, its not evaluable in elementary functions, try this one instead: \int_{- \infty}^{\infty} \frac{\ln(1+x^2)}{1+x^2} \ dx

Re: MX2 Integration Marathon Upon the substitution: x= \tan \theta The integral resolves into: \int \sin^2 \theta \ d\theta = \frac{1}{2} (\tan^{-1} x + \frac{x}{x^2+1}) + c ==== \int \frac{e^x}{\sqrt{(1+e^{2x})(1-e^{4x})}}} \ dx

Re: MX2 Integration Marathon Yep, nice. \int_0^1 \frac{\ln(1+x^2)}{1+x^2} \ dx

Re: MX2 Integration Marathon \int \sqrt{\csc x - \sin x} \ dx

Re: HSC 2013 4U Marathon Nice observation. So that can mean that when counting the probability the different all the cards needs to be on the table at once. And we need to find the probability of selecting the cards in that fashion among the m sets of (1,2,...,n) cards in one go (because...

Re: HSC 2013 4U Marathon Well yeah if m > n then we arrive at negative factorials so it should be discounted.

Re: HSC 2013 4U Marathon It becomes: \sum_{k=0}^{m } \frac{\binom{m}{k}}{\binom{n}{k}} And I don't know how to do that :/ Telescoping, pairing, binomial theorem don't seem to work Possibly a combinatoric proof? I suck at those but yeah EDIT: Considering the cards of numbers 1, 2, 3, 4...

Re: MX2 Integration Marathon Thanks. Integrals of that difficulty won't come up unless they actually give you the substitution (I'm pretty sure). But who knows maybe they will do something different this year.

Re: HSC 2013 3U Marathon Thread $Define the function$ \ \ I y = I(x) I \ \ $has the property that any real value input will output the integer closest to that value, in the case that it has decimal part$ \ \ .5 \ \ $it will take the number below it$ $i) Find$ \ \ I(\pi) \ \ $and$ \ \...

Re: MX2 Integration Marathon Divide by top and bottom by 1/x^3, by distributing the x's appropriately, we arrive at: \int \frac{(x+ 1/x +2)(1-1/x^2) \dx}{(x+1/x)^2 \sqrt{x^2 + 1 + 1/x^2}} dx Upon the substitution: y = x + 1/x The integral resolves into: \int \frac{y+2}{y^2 \sqrt{y^2-1}}...

About that one, I made it up on the spot without doing it, I intended the person to use the Gaussian pairing trick which can be used to solve the problem without the (-1)^k, I don't know if it works for this so if it doesn't I apologise lol I'll have a better look at it soon.

Re: HSC 2013 4U Marathon $Prove that$ $i)$ \ \ \binom{2n}{n} = \binom{n}{0}^2 + \binom{n}{1}^2 + \binom{n}{2}^2 + \dots + \binom{n}{n}^2 $ii) Find$ \ \ \sum_{k=0}^{n} (-1)^k k \binom{n}{k}^2

Re: HSC 2013 4U Marathon So is this how complex numbers are involved: (x-\alpha)^3 = \beta Is the minimal rational polynomial with roots alpha + cuberoot(beta). Following the same template as the square root conjugate proof we can prove that the roots of this polynomial above can...