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Assuming independence of the wheels and that each symbol is equally likely to show up, it is (5/20)*(4/20)*(2/20)*(2/20).

Re: Extracurricular Integration Marathon He used the Maclaurin series for the inverse hyperbolic sine (and the fact it holds at t = 1, which we can show for example by using Abel's theorem. We can show the series converges with the help of Stirling's approximation, for example).

$\noindent Let the common difference be $d \equiv a_{i+1} -a_{i}$. Note $\frac{1}{\sqrt{a_{i}} + \sqrt{a_{i +1}}} = \frac{\sqrt{a_{i+1}} -\sqrt{a_{i}}}{d}$, by ``rationalising the denominator'' and using $a_{i+1} -a_{i} = d$. The sum in the LHS of the equation to prove is thus $...

Use the reverse chain rule or substitution (let u equal the quadratic that appears).

$\noindent Hint: find the sum $S$ and product $P$ of the roots of the desired quadratic equation, then the desired equation will be $x^2 - Sx + P = 0$ (or any non-zero multiple of this).$

Re: HSC 2017 MX2 Integration Marathon $\noindent The integrand is $\sin t$, so yes, answer is $2$.$

Re: HSC 2017 MX2 Integration Marathon Which methods are you talking about? Inspection in general should be fine (might depend on wording of question), and you could differentiate your answer for indefinite integrals if you wanted to show 'working'.

Re: HSC 2017 MX2 Integration Marathon $\noindent Call that $N$-th partial product $P_{N}$, so with the hint, we have $P_{N} =\frac{\sin x}{2^{N}\sin \left( \frac{x}{2^{N}}\right)}$. Taking the limit as $N\to \infty$ shows that the infinite product is $\frac{\sin x}{x}$, using the fact that...

Re: HSC 2017 MX2 Integration Marathon $\noindent Also use $\lim_{\theta\to 0}\frac{\sin \theta}{\theta} = 1$.$

Re: HSC 2017 MX2 Integration Marathon See this document: https://www.math.ubc.ca/~feldman/m121/secx.pdf .

Re: HSC 2017 MX2 Integration Marathon Yeah, if you call it a trick.

Re: HSC 2017 MX2 Integration Marathon Here's a hint.

Re: HSC 2017 MX2 Integration Marathon Here's a much easier one. Find $$\int \sqrt{x^{2} + 4x + 8}\,\mathrm{d}x$$.

Re: HSC 2017 MX2 Integration Marathon Yep this is now integrable, but I don't think you can find a closed form answer for it using typical high school functions / concepts only (can do with the Sine integral, just by definition of the Sine integral. The curious student can read more about such...

Re: HSC 2017 MX2 Integration Marathon If you see a square in the denominator.

Re: HSC 2017 MX2 Integration Marathon It's basically just inspection. If you see some sum of product of functions like that, it might be a reverse product rule.

Re: HSC 2017 MX2 Integration Marathon $\noindent What question? When I say remove 2, something like this will converge: $\prod_{k=0}^{\infty} \cos\left(\frac{t}{2^{k}}\right)$. You can then integrate that (as in it will be integrable). It can be shown that that product converges to...

Re: HSC 2017 MX2 Integration Marathon I think this will be divergent. It will be convergent though if you remove the 2 from inside the product.

Re: HSC 2017 MX2 Integration Marathon $\noindent Using similar reversals of rules you can come up with other rules, like the reverse product rule: $\int \left(u'v +uv'\rght)\,\mathrm{d}x = uv + c$.$

Re: HSC 2017 MX2 Integration Marathon $\noindent It's that $\int \frac{u'v -uv'}{v^2}\,\mathrm{d}x = \frac{u}{v} +c$ (reversing the quotient rule). jathu123 inspected that the integrand was of the form $\frac{u'v -uv'}{v^2}$.$