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For those of us not in the known, could someone please enlighten us on some of these rumours about Dr Du. He's sounds like a very interesting character indeed.

Re: MX2 2016 Integration Marathon $\noindent Just trying to introduce a neat little method which avoids partial fractions to students who may not have seen such a thing before$

Re: MX2 2016 Integration Marathon $\noindent The question should read \textit{without} using polynomial long division and partial fractions.

Re: MX2 2016 Integration Marathon $\noindent \textbf{New Question}$ $\noindent Here is something a little easier. Find $\int \frac{x^4}{(2 - x)^3} \, dx$ without using polynomial long division and partial fractions.

Re: MX2 2016 Integration Marathon $\noindent Oh, now I see what you mean! Yes, you are right. Have corrected this in my solution. It was a damn typo.

Re: MX2 2016 Integration Marathon $\noindent I just added the term $-1 + 1$ as Paradoxica suggested when the question was posed.$

Re: MX2 2016 Integration Marathon $\noindent No, I believe everything is correct.$

Re: MX2 2016 Integration Marathon $\noindent (a) $\frac{d}{dx}(e^{f(x)} g(x) = e^{f(x)}[f'(x) g(x) + g'(x)]. $\noindent (b) To use the result of (a) in (b) we note that $f(x) = \cos x, f'(x) = -\sin x. $\noindent So for the term in square brackets we have$\\\begin{align*} f'(x) g(x) + g'(x)...

Re: Extracurricular Integration Marathon Yes, I know, but I was trying to keep the problem "purely real" for the benefit of any MX2 students who may care to read this thread (though those who are reading this thread probably already know or could follow the complex way anyway).

Re: Extracurricular Integration Marathon $\noindent We will evaluate a more general integral first. To do this we will make use of the following result$\\\sum^\infty_{k = 0} a^k \cos (kx) = \frac{1 - a \cos x}{1 - 2a \cos x + a^2}, \,\, |a| < 1. $\noindent To prove this result...

Re: Extracurricular Integration Marathon $\noindent We will make use of the following result$\\\sum^\infty_{k = 0} a^k \cos (kx) = \frac{1 - a \cos x}{1 - 2a \cos x + a^2}, \,\, |a| < 1.

Re: MX2 2016 Integration Marathon $\noindent I answered this question earlier in this thread in such a way that the problem with explicitly having to take limits is avoided. Also, I believe the answer should be positive.

Re: Extracurricular Integration Marathon $\noindent Do I smell a golden ratio? The whole $\sqrt{5}$ with 2 thing seems very suspicious to me.

Re: Extracurricular Integration Marathon $\noindent Okay, since the integrand is odd$ \begin{align*} \int^\infty_{-\infty} \sin(x^2) \, dx &= 2 \int^\infty_0 \sin (x^2) \, dx\end{align*} $\noindent Let $u = x^2, dx = \frac{du}{2 \sqrt{u}},$ while the limits of integration are...

Re: MX2 2016 Integration Marathon $\noindent Turning my previous question into something a little more interesting, suppose that$ I = \int \frac{1 + x^2}{1 - x^2} \frac{dx}{\sqrt{1 + x^4}}, \quad J = \int \frac{1 - x^2}{1 + x^2} \frac{dx}{\sqrt{1 + x^4}}. $\noindent By finding $I$ and...

Re: MX2 2016 Integration Marathon $\noindent Or for the first you could just use $u = x^2 + \frac{1}{x^2}$ straight off the bat. $\noindent The second question is a slight variation of a question asked in Spivak's \textit{Calculus} (3rd edition). In Q7 (v) on page 381 you are asked to...

Re: Extracurricular Integration Marathon $\noindent \textbf{Question No. 2} $\noindent Evaluate $\int^1_0 \frac{x^a - 1}{\ln x} \, dx$ where $a > 0.

Re: Extracurricular Integration Marathon $\noindent If we are allowed to use triple integrals, here is a slightly less familiar approach to this problem (the `standard' approach is one that typically uses a double integral). $\noindent Let $I = \int^\infty_{-\infty} e^{-x^2} \, dx.$ So...

Re: MX2 2016 Integration Marathon $\noindent Yes, here are two in the spirit of using a symmetric substitution as invoked in an earlier question of Paradoxica's. $\noindent By using symmetric substitutions, find $\noindent 1. $\int \frac{x^5 - x}{x^8 + 1} \, dx $\noindent 2. $\int...

Re: MX2 2016 Integration Marathon $\noindent Euler substitutions, no. But `coming up' with a clever substitution (which of course does not involve the hyperbolic functions), yes ;-)