HSC 2016 MX2 Integration Marathon (archive) (1 Viewer)

InteGrand · Mar 9, 2016

Re: Help i'm trapped in the title and can only speak once every 2 years

omegadot said:
$$\noindent Please let me know if there is a simpler way to do this one.$

$$\noindent Let $V = \int \tan^{-1} (2^{2n - 1} x^2) \, dx.$ Now let $2^{2n - 1} x^2 = \tan \theta, 2^{2n} x \, dx = \sec^2 \theta \, d\theta$ which after a little algebra becomes $dx = \frac{\sec^2 \theta}{2^{n + \frac{1}{2}} \sqrt{\tan \theta}} \, d\theta.$

$\begin{align*}V &= \frac{1}{2^{n + \frac{1}{2}}} \int \frac{\theta \sec^2 \theta}{\sqrt{\tan \theta}} \, d\theta\\&= \frac{1}{2^{n + \frac{1}{2}}} \theta \sqrt{\tan \theta} - \frac{1}{2^{n + \frac{1}{2}}} \int \sqrt{\tan \theta} \, d\theta + \frac{1}{2^{n + \frac{3}{2}}} \int \frac{\theta \sec^2 \theta}{\sqrt{\tan \theta}} \, d\theta \quad (\mbox{by parts})\\\Rightarrow V &= \frac{1}{2^{n - \frac{1}{2}}} \theta \sqrt{\tan \theta} -\frac{1}{2^{n - \frac{1}{2}}} \int \sqrt{\tan \theta} \, dx\end{align*}$

$$\noindent The integral $\int \sqrt{\tan x} \, dx$ can be found in a number of different ways. We will use the following approach: Let $I = \int \sqrt{\tan x} \, dx$ and $J = \int \sqrt{\cot x} \, dx.$

$\begin{align*}I + J &= \int \left (\sqrt{\tan x} + \sqrt{\cot x} \right ) \, dx\\&=\int \frac{\sin x + \cos x}{\sqrt{\sin x \cos x}} \, dx\\&= \sqrt{2} \int \frac{\sin x + \cos x}{\sqrt{2 \sin x \cos x}} \, dx\\&= \sqrt{2} \int \frac{\sin x + \cos x}{\sqrt{1 + 2 \sin x \cos x - (\cos^2 x + \sin^2 x)}} \, dx\\&= \sqrt{2} \int \frac{\sin x + \cos x}{\sqrt{1 - (\sin^2 x - 2 \sin x \cos x + \cos^2 x)}} \, dx\\&= \sqrt{2} \int \frac{\sin x + \cos x}{\sqrt{1 - (\sin x - \cos x)^2}} \, dx\\&= \sqrt{2} \sin^{-1} (\sin x - \cos x) + {\cal{C}}_1\quad (*)\end{align*}$

$$\noindent And\\\begin{align*}I - J &= \int \left (\sqrt{\tan x} - \sqrt{\cot x} \right ) \, dx\\&= \int \frac{\sin x - \cos x}{\sqrt{\sin x \cos x}} \, dx\\&= \sqrt{2} \int \frac{\sin x - \cos x}{\sqrt{2 \sin x \cos x}} \, dx\\&= \sqrt{2} \int \frac{\sin x - \cos x}{\sqrt{(\cos^2 x + \sin^2 x) + 2\sin x \cos x - 1}} \, dx\\&= \sqrt{2} \int \frac{\sin x - \cos x}{\sqrt{(\cos x + \sin x)^2 - 1}} \, dx\\&= -\sqrt{2} \ln \left |\sqrt{(\cos x + \sin x)^2 - 1} + \sin x + \cos x \right | + {\cal{C}}_2\\ &= -\sqrt{2} \ln \left |\sqrt{\sin 2x} + \sin x + \cos x \right | + {\cal{C}}_2 \quad (**)\end{align*}$

$\noindent Adding$ (*)$ and $(**)$ gives$

$2 I = \sqrt{2} \sin^{-1} (\sin x - \cos x) - \sqrt{2} \ln \left |\sqrt{\sin 2x} + \sin x + \cos x \right | + {\cal{C}}$

$\noindent or$

$\begin{align*}\int \sqrt{\tan x} \, dx &= \frac{1}{\sqrt{2}} \sin^{-1} (\sin x - \cos x) - \frac{1}{\sqrt{2}} \ln \left |\sqrt{\sin 2x} + \sin x + \cos x \right | + \cal{C}\end{align*}$

$$\noindent Thus$\\\begin{align*}V &= \frac{1}{2^{n - \frac{1}{2}}} \theta \sqrt{\tan \theta} - \frac{1}{2^n} \sin^{-1} (\sin \theta - \cos \theta) - \frac{1}{2^n} \ln \left |\sqrt{\sin 2\theta} + \sin \theta + \cos \theta \right | + \cal{C}\end{align*}$

$$\noindent Noting that $\sin \theta = \frac{2^{2n - 1} x^2}{\sqrt{2^{4n - 2} x^4 + 1}}$ and $\cos \theta = \frac{1}{\sqrt{2^{4n - 2} + 1}}$. Substituting into the expression for $V$, after some algebra, we finally have$\\ \begin{align*}\int \tan^{-1} (2^{2n - 1} x^2) \, dx &= x \tan^{-1} (2^{2n - 1} x^2) - \frac{1}{2^n} \sin^{-1} \left (\frac{2^{2n - 1} x^2 - 1}{\sqrt{2^{4n - 2} x^4 + 1}} \right ) + \frac{1}{2^n} \ln \left |\frac{2^{2n - 1} x^2 + 2^n x + 1}{\sqrt{2^{4n - 2} x^4 + 1}} \right | + \cal{C}.\end{align*}$

Why did you write ''Re: Help i'm trapped in the title and can only speak once every 2 years'' in the reply title thing? Just for fun?

Edit: oh, it's because Paradoxica had written ''Help i'm trapped in the title and can only speak once every 2 years'' in his integral post.

omegadot · Mar 10, 2016

Re: MX2 2016 Integration Marathon

juantheron said:
$Hence show that $\int_{0}^{1}x^k\left(1-x\right)^{n-k}dx = \left[\binom{n}{k}(n+1)\right]^{-1}$ for $k=0,1,2,3,.....,n$

$\noindent I will show the second result directly. Let$

$I_k = \int^1_0 x^k (1 - x)^{n - k} \, dx.$

$$\noindent Now $I_0 = \int^1_0 (1 - x)^n \, dx = \int^1_0 x^n \, dx = \frac{1}{n + 1}.$

$\begin{align*}I_k &= \left [\frac{x^{k + 1}}{k + 1} (1 - x)^{n - k} \right ]^1_0 + \frac{n - k}{k + 1} \int^1_0 x^{k + 1} (1 - x)^{n - (k + 1)} \, dx \quad \mbox{(by parts)}\\&= \frac{n - k}{k + 1} I_{k + 1}\\\Rightarrow I_k + 1 &= \frac{k + 1}{n - k}I_k, \quad k = 0,1,2,\ldots,n - 1\end{align*}$

$$\noindent or $I_k = \frac{k}{n - k + 1} I_{k - 1}$ for $k = 0,1,2,\ldots,n$ on replacing $k$ with $k - 1$.$

$\noindent Inductively on $k$ we have$

$\begin{align*}I_k &= \frac{k}{n - k + 1} \cdot \frac{k - 1}{n - k + 2} \cdots \frac{2}{n - 1} \cdot \frac{1}{n} \cdot I_0\\&= \frac{k!}{(n - k + 1)(n - k + 2) \cdots (n - 1) n} \cdot I_0\\&= \frac{k! (n - k) \cdots 2 \cdot 1}{n (n - 1) \cdots (n - k + 1)(n - k) \cdots 2 \cdot 1}\cdot I_0\\&= \frac{k! (n - k)!}{n!} \cdot I_0\\&= \frac{k! (n - k)!}{n!} \cdot \frac{1}{n + 1}\\&= \frac{1}{\binom{n}{k}} \cdot \frac{1}{n + 1}\\&= \left [\binom{n}{k} (n + 1) \right ]^{-1},\end{align*}$

$$\noindent as required to show.$

omegadot · Mar 10, 2016

Re: MX2 2016 Integration Marathon

$$\noindent \textbf{Next Question}$

$$\noindent Find $\int \frac{\sin x + \sin 2x + \sin 3x + \sin 4x}{\cos x + \cos 2x + \cos 3x + \cos 4x} \, dx$

Paradoxica · Mar 10, 2016

Re: MX2 2016 Integration Marathon

omegadot said:
$$\noindent \textbf{Next Question}$

$$\noindent Find $\int \frac{\sin x + \sin 2x + \sin 3x + \sin 4x}{\cos x + \cos 2x + \cos 3x + \cos 4x} \, $d$x$

$$\noindent Consider the general case of the integrand.$\\ \int \frac{\sum_{r=1}^{n}\sin{rx}}{\sum_{r=1}^{n}\cos{rx}}\,$d$x = \int \frac{\Im{(\sum_{r=1}^{n} e^{r i x})}}{\Re{(\sum_{r=1}^{n} e^{r i x})}} \, $d$x\\ \sum_{r=1}^{n} e^{r i x} = \frac{e^{ix}(e^{n i x}-1)}{e^{ix}-1}\\$Using the easily derivable result $e^{i \alpha}-1 \equiv 2i\sin{\frac{\alpha}{2}}e^{\frac{i\alpha}{2}}$, we have:$\\ \frac{e^{ix}(e^{n i x}-1)}{e^{ix}-1} \equiv e^{ix} \frac{2i\sin{\frac{nx}{2} e^{\frac{n i x}{2}}}}{2i \sin{\frac{x}{2}} e^{\frac{ix}{2}}} \equiv \frac{\sin{\frac{nx}{2}}}{\sin{\frac{x}{2}}} e^{\frac{(n+1)ix}{2}} \equiv \lambda (\cos{\frac{(n+1)x}{2}}+i\sin{\frac{(n+1)x}{2}}) \\ \frac{\Im{(\sum_{r=1}^{n} e^{r i x})}}{\Re{(\sum_{r=1}^{n} e^{r i x})}} \equiv \tan{\frac{(n+1)x}{2}} \\ \int \frac{\Im{(\sum_{r=1}^{n} e^{r i x})}}{\Re{(\sum_{r=1}^{n} e^{r i x})}} \,$d$x = \frac{2}{n+1}\log{\left (\sec{\frac{(n+1)x}{2}}\right )} +\cal{C}$

$$\noindent Setting $n=4$, we have $\\ \int \frac{\sin x + \sin 2x + \sin 3x + \sin 4x}{\cos x + \cos 2x + \cos 3x + \cos 4x} \, $d$x = \frac{2}{5} \log{\left( \sec{\frac{5x}{2}} \right)}+\cal{C}$

InteGrand · Mar 10, 2016

Re: MX2 2016 Integration Marathon

leehuan said:
$\int{dx}$

Lol ok I'm sorry.

$x+c$

omegadot · Mar 10, 2016

Re: MX2 2016 Integration Marathon

Paradoxica said:
$$\noindent Consider the general case of the integrand.$\\ \int \frac{\sum_{r=1}^{n}\sin{rx}}{\sum_{r=1}^{n}\cos{rx}}\,$d$x = \int \frac{\Im{(\sum_{r=1}^{n} e^{r i x})}}{\Re{(\sum_{r=1}^{n} e^{r i x})}} \, $d$x\\ \sum_{r=1}^{n} e^{r i x} = \frac{e^{ix}(e^{n i x}-1)}{e^{ix}-1}\\$Using the easily derivable result $e^{i \alpha}-1 \equiv 2i\sin{\frac{\alpha}{2}}e^{\frac{i\alpha}{2}}$, we have:$\\ \frac{e^{ix}(e^{n i x}-1)}{e^{ix}-1} \equiv e^{ix} \frac{2i\sin{\frac{nx}{2} e^{\frac{n i x}{2}}}}{2i \sin{\frac{x}{2}} e^{\frac{ix}{2}}} \equiv \frac{\sin{\frac{nx}{2}}}{\sin{\frac{x}{2}}} e^{\frac{(n+1)ix}{2}} \equiv \lambda (\cos{\frac{(n+1)x}{2}}+i\sin{\frac{(n+1)x}{2}}) \\ \frac{\Im{(\sum_{r=1}^{n} e^{r i x})}}{\Re{(\sum_{r=1}^{n} e^{r i x})}} \equiv \tan{\frac{(n+1)x}{2}} \\ \int \frac{\Im{(\sum_{r=1}^{n} e^{r i x})}}{\Re{(\sum_{r=1}^{n} e^{r i x})}} \,$d$x = \frac{2}{n+1}\log{\left (\sec{\frac{(n+1)x}{2}}\right )} +\cal{C}$

$$\noindent Setting $n=4$, we have $\\ \int \frac{\sin x + \sin 2x + \sin 3x + \sin 4x}{\cos x + \cos 2x + \cos 3x + \cos 4x} \, $d$x = \frac{2}{5} \log{\left( \sec{\frac{5x}{2}} \right)}+\cal{C}$

$$\noindent For those who wish to keep it purely real, various product-to-sum and sum-to-product identities for the trig functions can be used to find each of the sums. For the general case we will consider $\frac{\sum^{n - 1}_{k = 1} \sin k\theta}{\sum^{n - 1}_{k = 1} \cos k\theta}.$

$\noindent For the numerator, multiplying it by the term $2\sin \frac{\theta}{2}$ we have$

$\begin{align*}2\sin \frac{\theta}{2} \sum^{n - 1}_{k = 1} \sin k\theta = 2 \sin \frac{\theta}{2} \sin \theta + 2 \sin \frac{\theta}{2} \sin 2\theta + \cdots + 2 \sin \frac{\theta}{2} \sin (n -2)\theta + 2 \sin \frac{\theta}{2} \sin (n - 1)\theta.\end{align*}$

$\noindent On applying the product-to-sum identity of $2 \sin A \sin B = \cos (A - B) - \cos (A + B)$ to each term we have$

$\begin{align*}2 \sin \frac{\theta}{2} \sin \theta &= \cos \frac{\theta}{2} - \cos \frac{3\theta}{2}\\2 \sin \frac{\theta}{2} \sin 2\theta &= \cos \theta \frac{3\theta}{2} - \cos \frac{5\theta}{2}\\\quad \vdots &\\2 \sin \frac{\theta}{2} \sin (n - 2)\theta &= \cos (n - \frac{5}{2})\theta - \cos (n - \frac{3}{2}) \theta\\2 \sin \frac{\theta}{2} \sin (n - 1)\theta &= \cos (n - \frac{3}{2})\theta - \cos (n - \frac{1}{2})\theta\end{align*}$

$\noindent On summing the terms, the series ``telescopes'' with all but the first and last terms cancelling out, leaving$

$\begin{align*}2 \sin \frac{\theta}{2} \sum^{n - 1}_{k = 1} \sin k\theta &= \cos \frac{\theta}{2} - \cos (n - \frac{1}{2}) \theta = 2 \sin \frac{n\theta}{2} \sin \left (\frac{n - 1}{2} \right )\theta.\end{align*}$

$$\noindent Here the sum-to-product rule of $\cos A - \cos B = - 2 \sin \left (\frac{A+B}{2} \right ) \sin \left (\frac{A - B}{2} \right )$ has been applied. Thus$\sum^{n - 1}_{k = 1} \sin k\theta = \frac{\sin \frac{n\theta}{2} \sin \left (\frac{n - 1}{2} \right )\theta}{\sin \frac{\theta}{2}}.$

$\noindent For the denominator, first multiply it by $2 \sin \frac{\theta}{2}$ then in a similar fashion as to what was done for the numerator, it can be readily shown that$

$\begin{align*} \sum^{n - 1}_{k = 1} \cos k\theta &= \frac{\cos \frac{n\theta}{2} \sin \left (\frac{n - 1}{2} \right )\theta}{\sin \frac{\theta}{2}}.\end{align*}$

$\noindent Thus we see that$

$\begin{align*}\frac{\sum^{n - 1}_{k = 1} \sin k\theta}{\sum^{n - 1}_{k = 1} \cos k\theta} = \frac{\sin \theta + \sin 2\theta + \cdots + \sin (n - 1)\theta}{\cos \theta + \cos 2\theta + \cdots + \cos (n - 1)\theta} = \tan \frac{n\theta}{2}.\end{align*}$

Paradoxica · Mar 10, 2016

Re: MX2 2016 Integration Marathon

$\int \frac{\sum_{r=0}^{n}\sin{rx}}{\sum_{r=0}^{n}\cos{rx}}\,$d$x$

agha · Mar 10, 2016

Re: MX2 2016 Integration Marathon

$\\ \lim_{a \to \infty} \int_0^a \frac{\text{d}x}{(1+x^2)(1+x^3)}$

Drsoccerball · Mar 10, 2016

Re: MX2 2016 Integration Marathon

agha said:
$\\ \lim_{a \to \infty} \int_0^a \frac{\text{d}x}{(1+x^2)(1+x^3)}$

$$ You do the shifty $ x = \frac{1}{u} $ substitution and add the two dummy variables to end up with $ : \\ 2I=\int_0^{\infty}{\frac{1}{x^2+1}}dx$

$I=\frac{\pi}{4} $ Sorry for crappy solution im very busy $$

agha · Mar 10, 2016

Re: MX2 2016 Integration Marathon

Drsoccerball said:
$$ You do the shifty $ x = \frac{1}{u} $ substitution and add the two dummy variables to end up with $ : \\ 2I=\int_0^{\infty}{\frac{1}{x^2+1}}dx$

$I=\frac{\pi}{4} $ Sorry for crappy solution im very busy $$

Yes well done, alternatively $x= \tan u$ and then $v = \frac{\pi}{2} - u$ , adding up the integrals in u and v to get what you have.

agha · Mar 10, 2016

Re: MX2 2016 Integration Marathon

$\\ $Evaluate$ \ \frac{\int_0^1 (1-x^n)^m \ dx}{\int_0^1 (1-x^n)^{m+1} \ dx} \ $for positive integers$ \ m,n$

Paradoxica · Mar 10, 2016

Re: MX2 2016 Integration Marathon

agha said:
$\\ $Evaluate$ \ \frac{\int_0^1 (1-x^n)^m \ dx}{\int_0^1 (1-x^n)^{m+1} \ dx} \ $for positive integers$ \ m,n$

$$\noindent Using reduction, $\frac{(m+1)n+1}{(m+1)n}I_{m+1}=I_m\\ $Hence, the ratio is equal to $1+\frac{1}{(m+1)n}$

Paradoxica · Mar 11, 2016

Re: MX2 2016 Integration Marathon

omegadot said:
$$\noindent \textbf{Next Question}$

$$\noindent Find $ \int \frac{(2 - x^2) e^x}{(1 - x) \sqrt{1 - x^2}} \, dx.$

A simple reversal of the product rule does away with the integral. The hard part is figuring out how to split the algebraic part.

Paradoxica · Mar 11, 2016

Re: Help i'm trapped in the title and can only speak once every 2 years

Paradoxica said:
$$\noindent$ \int 20n x^{5n-1} \tan^{-1}\left( \frac{(x^n-1)x^n(x^n+1)}{(x^{2n}-\sqrt{3}x^n+1)(x^{2n}+\sqrt{3}x^n+1)} \right)\,$d$x$

Nobody has touched this yet. Will do soon if untouched for a bit longer.

omegadot · Mar 11, 2016

Re: Help i'm trapped in the title and can only speak once every 2 years

Paradoxica said:
$$\noindent$ \int 20n x^{5n-1} \tan^{-1}\left( \frac{(x^n-1)x^n(x^n+1)}{(x^{2n}-\sqrt{3}x^n+1)(x^{2n}+\sqrt{3}x^n+1)} \right)\,$d$x$

$$\noindent I will assume $n \neq 0.$

$\begin{align*}I &= \int 20n x^{5n-1} \tan^{-1}\left( \frac{(x^n-1)x^n(x^n+1)}{(x^{2n}-\sqrt{3}x^n+1)(x^{2n}+\sqrt{3}x^n+1)} \right)\,dx\\&= \int 20 n x^{5n - 1} \tan^{-1} \left (\frac{x^n (x^{2n} - 1)}{x^{4n} - x^{2n} + 1} \right ) \, dx\\ &= 4 x^{5n} \tan^{-1} \left (\frac{x^n (x^{2n} - 1)}{x^{4n} - x^{2n} + 1} \right ) - \int 4 x^{5n} \cdot \frac{d}{dx} \left [\tan^{-1} \left (\frac{x^n (x^{2n} - 1)}{x^{4n} - x^{2n} + 1} \right ) \right ] \, dx \quad \mbox{(by parts)}\end{align*}$

$\noindent Now$

$\begin{align*}\frac{d}{dx} \left [\tan^{-1} \left (\frac{x^n (x^{2n} - 1)}{x^{4n} - x^{2n} + 1} \right ) \right ] &= -\frac{n x^{n - 1} (x^{6n} - 2x^{4n} - 2x^{2n} + 1)}{x^{8n} - x^{6n} + x^{4n} - x^{2n} + 1}\end{align*}$

$\noindent Thus$

$\begin{align*}I &= 4 x^{5n} \tan^{-1} \left (\frac{x^n (x^{2n} - 1)}{x^{4n} - x^{2n} + 1} \right ) +4n \int \frac{x^{6n - 1} (x^{6n}-2x^{4n} - 2x^{2n} + 1)}{x^{8n} - x^{6n} + x^{4n} - x^{2n} + 1} \, dx\end{align*}$

$\noindent Now considering this last integral. Calling it $J$ we have$

$\begin{align*}J &= \int \frac{x^{6n - 1} (x^{6n}-2x^{4n} - 2x^{2n} + 1)}{x^{8n} - x^{6n} + x^{4n} - x^{2n} + 1} \, dx = \int \frac{x^{2n - 1} (x^{2n})^2 [(x^{2n})^3 - 2(x^{2n})^2 - 2x^{2n} + 1]}{(x^{2n})^4 - (x^{2n})^3 + (x^{2n})^2 - x^{2n} + 1} \, dx\end{align*}$

$\noindent Now let $u = x^{2n}, du = 2n x^{2n - 1} \, dx \Rightarrow x^{2n - 1} \, dx = \frac{du}{2n}$. Thus$

$\begin{align*}2nJ &= \int \frac{u^2(u^3 - 2u^2 - 2u + 1)}{u^4 - u^3 + u^2 - u + 1} \, du\\&= \int \frac{u (u^4 - 2u^3 - 2u^2 + u)}{u^4 - u^3 + u^2 - u + 1} \, dx\\&= \int \frac{u[(u^4 - u^3 + u^2 - u + 1)-u^3 - 3u^2 + 2u - 1]}{u^4 - u^3 + u^2 - u + 1} \, du\\&= \int u \, du - \int \frac{u^4 + 3u^3 - 2u^2 + u}{u^4 - u^3 + u^2 - u + 1} \, du\\&= \frac{u^2}{2} - \int \frac{(u^4 - u^3 + u^2 - u + 1)+4u^3 - 3u^2 + 2u - 1}{u^4 - u^3 + u^2 - u + 1} \, du\\&= \frac{u^2}{2} - \int du - \int \frac{4u^3 - 3u^2 + 2u - 1}{u^4 - u^3 + u^2 - u + 1} \, du\\&= \frac{u^2}{2} - u - \ln \left |u^4 - u^3 + u^2 - u + 1 \right | + \cal{C}\end{align*}$

$$\noindent So finally, on substituting $x^{2n}$ for $u$ into $J$, we have for $I$

$\begin{align*}I &= 4 x^{5n} \tan^{-1} \left (\frac{x^n (x^{2n} - 1)}{x^{4n} - x^{2n} + 1} \right ) + x^{4n} - 2 x^{2n} - 2 \ln \left |x^{8n} - x^{6n} + x^{4n} - x^{2n} + 1 \right | + \cal{C}.\end{align*}$

InteGrand · Mar 11, 2016

Re: MX2 2016 Integration Marathon

$\noindent Let $F_1 (x) = \int \ln (x) \text{ d}x$, and for integers $n\geq 2$, let $F_n (x) = \int F_{n-1} (x) \text{ d}x$. Find a formula for $F_n (x)$ involving $n$. (Your answer may be expressed with the help of Sigma notation, and should contain $n$ arbitrary constants.)$

KingOfActing · Mar 11, 2016

Re: MX2 2016 Integration Marathon

InteGrand said:
$\noindent Let $F_1 (x) = \int \ln (x) \text{ d}x$, and for integers $n\geq 2$, let $F_n (x) = \int F_{n-1} (x) \text{ d}x$. Find a formula for $F_n (x)$ involving $n$. (Your answer may be expressed with the help of Sigma notation, and should contain $n$ arbitrary constants.)$

$$By manually finding the first few terms, it becomes clear that $ F_n(x) = \frac{x^n}{n!}\ln{x} - C_nx^n + P_{n-1}(x) $ where $ C_n $ represents a constant dependent on $n$ and $ P_n(x) $ is an $n$th degree polynomial with arbitrary coefficients. Using the given formula and one application of integration by parts, it can be shown that $ C_n $ satisfies the recurrence relation $ C_n = \frac{C_{n-1}}{n} + \frac{1}{n \times n!} $ with initial condition $ C_1 = 1 $. It can be proven by induction that $C_n = \frac{\sum_{m=1}^{n}\frac{1}{m}}{n!}$

Paradoxica · Mar 12, 2016

Pi is rational!!! Don't fall for the lies of the government!!!!

$\int \frac{2(x+2)(x^4+1)}{x^8-2x^4+1}\,$d$x$

Paradoxica · Mar 12, 2016

e is algebraic!!!! Don't fall for the lies of the government!!!!!

For those who aren't bothered doing the above integral, here is an intriguing generalised integral.
$\int_{-a}^a \frac{\text{d}x}{b+x^{2 n-1}+\sqrt{b^2+x^{4 n-2}}}$
Remember to check special cases.

Paradoxica · Mar 29, 2016

Re: MX2 2016 Integration Marathon

$$\noindent Given $n$ is a non-negative integer, find the recurrence relation for $\\ I_n = \int_{-\infty}^\infty \frac{\text{d}x}{(x^2+1)^{n+1}} \\ $Hence, prove the following statement using induction:$ \\ \int_{-\infty}^\infty \frac{\text{d}x}{(x^2+1)^{n+1}} = \frac{(2n)!\,\pi}{2^{2n}(n!)^2}$

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