HSC 2016 MX2 Integration Marathon (archive) (4 Viewers)

KingOfActing · Nov 24, 2015

Re: MX2 2016 Integration Marathon

leehuan said:
Get over it Paradox. When it comes to maths you're at latest a 15'er

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Not sure if KingOfActing will go for it

I just tried it but I think I made a mistake somewhere

$$Start with the substitution $ x + 1 = \sqrt{6}\tan{\theta} $ So after simplification the integral becomes $ \int 6\tan^2{\theta}\sec{\theta} -5\sqrt{6}\tan{\theta}\sec{\theta} - 3\sec{\theta} \, d\theta$

After heaps of integration by parts our integral becomes $3\tan{\theta}\sec{\theta} -6\ln{|\sec{\theta} + \tan{\theta}|} - 5\sqrt{6}\sec{\theta} + C$

Then back substitution using $\tan{\theta} = \frac{x+1}{\sqrt{6}} $ and $ \sec{\theta} = \frac{\sqrt{x^2+2x+7}}{\sqrt{6}}$ and lots of simplification, our integral can be written as $\frac{1}{2}(x-9)\sqrt{x^2+2x+7} - 6\ln{|x+1 +\sqrt{x^2+2x+7}|} + C$

InteGrand · Nov 24, 2015

Re: MX2 2016 Integration Marathon

$\underline{NEW QUESTION}$

$Find $\int \sqrt{x + \sqrt{x^2 + 1}} \text{ d}x$.$

porcupinetree · Nov 24, 2015

Re: MX2 2016 Integration Marathon

InteGrand said:
$\underline{NEW QUESTION}$

$Find $\int \sqrt{x + \sqrt{x^2 + 1}} \text{ d}x$.$

I think I've got it, let me write up a solution

omegadot · Nov 25, 2015

Re: MX2 2016 Integration Marathon

porcupinetree said:
Next question:

$\int \frac{x^{2} - 7x + 3}{\sqrt{x^{2} + 2x + 7}} \, dx$

$\noindent There is probably a more elegant way to do this, and it is a bit of a slog fest, but this is what I did. I started by rewriting the integrand as follows:$

$\begin{align*}\int \frac{x^2 - 7x + 3}{\sqrt{x^2 + 2x + 7}} \, dx&=\int \frac{(x + 1)^2 - 9 (x + 1) + 11}{\sqrt{(x + 1)^2 + 6}} dx\\&= \int \frac{(x + 1)^2}{\sqrt{(x + 1)^2 + 6}} \, dx - 9\int \frac{x + 1}{\sqrt{(x+1)^2 + 6}} \, dx + 11 \int \frac{dx}{\sqrt{(x + 1)^2 + 6}} \, dx\\&= I_1 - 9 I_2 + 11 I_3\end{align*}$

$$\noindent Using the result $\int \frac{du}{\sqrt{u^2 + a^2}} \, du = \ln (u + \sqrt{u^2 + a^2}) + {\cal{C}}$ we have$\\\begin{align*}I_3 = \ln \left ((x+1) +\sqrt{(x+1)^2 + (\sqrt{6})^2} \right ) + {\cal{C}}_3\end{align*}$

$$\noindent For $I_2$ using the substitution $u = (x + 1)^2 + 6, du = 2(x + 1) \, dx$ one has$\\\begin{align*}I_2 &= \frac{1}{2} \int \frac{du}{\sqrt{u}} =\sqrt{u} + {\cal{C}}_2 =\sqrt{(x + 1)^2 + 6} + {\cal{C}}_2\end{align*}$

$$\noindent For $I_1$ using the substitution $x + 1 = \sqrt{6} \tan \theta, dx = \sqrt{6} \sec^2 \theta \, d\theta$ we have\\\begin{align*}I_1 &= 6 \int \tan^2 \theta \sec \theta \, d\theta\\&= 6 \int (\sec^2 \theta - 1) \sec \theta \, d\theta\\&= 6 \int \sec^3 \theta \, d\theta - 6 \int \sec \theta \, d\theta\\&= 6 \cdot \frac{1}{2} \sec \theta \tan \theta + 6 \cdot \frac{1}{2} \ln |\sec \theta + \tan \theta | - 6 \ln |\sec \theta + \tan \theta| + {\cal{C}}_1\\&= 3 \sec \theta \tan \theta - 3 \ln| \sec \theta + \tan \theta| + {\cal{C}}_1\end{align*}$

$$\noindent Note the well-known results for the indefinite integrals of secant and secant cube have been used here. Now since $\tan \theta = \frac{x + 1}{\sqrt{6}} \Rightarrow \sec \theta = \frac{\sqrt{(x + 1)^2 + 6}}{\sqrt{6}}$ so one has$\\\begin{align*}I_1 &= \frac{1}{2} (x + 1) \sqrt{(x + 1)^2 + 6} - 3 \ln \left ((x + 1) + \sqrt{(x + 1)^2 + 6} \right ) + {\cal{C}}_1\end{align*}$ giving a final result of$\\\begin{align*} \int \frac{x^2 - 7x + 3}{\sqrt{x^2 + 2x + 7}} \, dx &= \frac{1}{2} (x - 17) \sqrt{x^2 + 2x + 7} + 8 \ln \left (x + 1 + \sqrt{x^2 + 2x + 7} \right ) + {\cal{C}}\end{align*}$

porcupinetree · Nov 25, 2015

Re: MX2 2016 Integration Marathon

porcupinetree said:
I think I've got it, let me write up a solution

There are probably several silly errors in here somewhere, keep in mind that my brain has been in holiday mode for about a month

omegadot · Nov 25, 2015

Re: MX2 2016 Integration Marathon

$$\noindent \textbf{Next Question} -- something a little easier$

$$\noindent Find $ \int \cos^{12}(5x) \sin^3 (5x) \, dx.$

dan964 · Nov 25, 2015

Re: MX2 2016 Integration Marathon

omegadot said:
$$\noindent \textbf{Next Question} -- something a little easier$

$$\noindent Find $ \int \cos^{12}(5x) \sin^3 (5x) \, dx.$

$\int \cos^{12}(5x) \sin (5x) \times (1-\cos^2 (5x)) \, dx.$
$\int \cos^{12}(5x) \sin (5x) - cos^{14}(5x) \sin (5x) \, dx.$
$= -\frac{1}{65} cos^{13}(5x) + \frac{1}{75} cos^{15}(5x) + C.$

omegadot · Nov 25, 2015

Re: MX2 2016 Integration Marathon

$\noindent \textbf{Next Question}$

$$\noindent Find $\int \frac{\sqrt[4]{1 + \sqrt[3]{x}}}{x} \, dx$

leehuan · Nov 25, 2015

Re: MX2 2016 Integration Marathon

Sent from my iPhone using Tapatalk

porcupinetree · Nov 25, 2015

Re: MX2 2016 Integration Marathon

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

$$\noindent Find $\int \frac{\sqrt[4]{1 + \sqrt[3]{x}}}{x} \, dx$

EDIT: whoops didn't see that leehuan had already done it

omegadot · Nov 25, 2015

Re: MX2 2016 Integration Marathon

porcupinetree said:
EDIT: whoops didn't see that leehuan had already done it

$$\noindent Yes but each of your approaches were slightly different and leehuan accidently dropped a factor of 12 in the first term of the final answer.$

omegadot · Nov 25, 2015

Re: MX2 2016 Integration Marathon

$\noindent \textbf{Next Question}$

$$\noindent Find $ \int \left (\frac{1 - \sin x}{1 - \cos x} \right ) e^x \, dx$

leehuan · Nov 25, 2015

Re: MX2 2016 Integration Marathon

omegadot said:
$$\noindent Yes but each of your approaches were slightly different and leehuan accidently dropped a factor of 12 in the first term of the final answer.$

I realised. Whoops.

Ekman · Nov 25, 2015

Re: MX2 2016 Integration Marathon

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

$$\noindent Find $ \int \left (\frac{1 - \sin x}{1 - \cos x} \right ) e^x \, dx$

$\frac{d}{dx} \left( \frac{e^x \cdot \sin{x}}{1-\cos{x}} \right) = \frac{e^x(\sin{x} - 1)}{1-\cos{x}}$

$\therefore \int \left (\frac{1 - \sin x}{1 - \cos x} \right ) e^x \, dx = - \left( \frac{e^x \cdot \sin{x}}{1-\cos{x}} \right) + c$

(This question reminds me of the last question of the 2014 hsc, where doing IBP, via inspection can solve it in a couple of steps)

InteGrand · Nov 25, 2015

Re: MX2 2016 Integration Marathon

$\textbf{NEW QUESTION}$

$Define $f_1 (x) = |x|$ and for integers $n\geq 2$, $f_n (x) = \Big{|} x + f_{n-1} (x)\Big{|}$.$

$Let $\mathcal{I}_n =\int _{-1} ^{1} f_n (x) \text{ d}x$, for $n= 1,2,3,\ldots$.$

$Derive an expression for $\mathcal{I}_n$ in terms of $n$.$

KingOfActing · Nov 25, 2015

Re: MX2 2016 Integration Marathon

InteGrand said:
$\textbf{NEW QUESTION}$

$Define $f_1 (x) = |x|$ and for integers $n\geq 2$, $f_n (x) = \Big{|} x + f_{n-1} (x)\Big{|}$.$

$Let $\mathcal{I}_n =\int _{-1} ^{1} f_n (x) \text{ d}x$, for $n= 1,2,3,\ldots$.$

$Derive an expression for $\mathcal{I}_n$ in terms of $n$.$

$\textbf{Case 1: } x \geq 0\\$Proposition: $ f_n(x) = nx\\$Proof: $ f_1(x) = |x| = x\\$Assume for $ n = k $ : $ f_k(x) = kx\\f_{k+1}(x) = |x+f_k(x)| = |x+kx| = (k+1)x\\$Hence by the principle of mathematical induction our proposition has been proven.$\\\\\textbf{Case 2: } x < 0\\\textbf{Case 2.1: } n = 2m,\; m \in \mathbb{N}\\$Proposition: $ f_n(x) = 0\\$Proof: $ f_2(x) = |x+f_1(x)| = |x+|x|| = |x-x| = 0\\$Assume for $ n = 2k $ : $ f_{2k}(x) = 0\\f_{2k+2}(x) = |x+f{2k+1}(x)| = |x+|x+f_{2k}(x)|| = |x+|x+0|| = |x+|x|| = |x-x| = 0\\$Hence by the principle of mathemtical induction our prposition has been proven.$\\\textbf{Case 2.2: } n = 2m + 1, \; m \in \mathbb{N}\\$Proposition: $ f_n(x) = -x\\$Proof: $ f_{2k+1}(x) = |x+f_{2k}(x)| = |x+0| = |x| = -x\\\\ \therefore \mathcal{I}_n = \int_{-1}^{1}f_n(x)\,dx = \int_{-1}^{0} f_n(x)\,dx + \int_{0}^{1} f_n(x)\,dx = \frac{n}{2} $ in the case that $n$ is even, or $ \frac{n+1}{2} $ in the case that n is odd.$$

Sorry my formatting isn't the best

InteGrand · Nov 25, 2015

Re: MX2 2016 Integration Marathon

KingOfActing said:
$\textbf{Case 1: } x \geq 0\\$Proposition: $ f_n(x) = nx\\$Proof: $ f_1(x) = |x| = x\\$Assume for $ n = k $ : $ f_k(x) = kx\\f_{k+1}(x) = |x+f_k(x)| = |x+kx| = (k+1)x\\$Hence by the principle of mathematical induction our proposition has been proven.$\\\\\textbf{Case 2: } x < 0\\\textbf{Case 2.1: } n = 2m,\; m \in \mathbb{N}\\$Proposition: $ f_n(x) = 0\\$Proof: $ f_2(x) = |x+f_1(x)| = |x+|x|| = |x-x| = 0\\$Assume for $ n = 2k $ : $ f_{2k}(x) = 0\\f_{2k+2}(x) = |x+f{2k+1}(x)| = |x+|x+f_{2k}(x)|| = |x+|x+0|| = |x+|x|| = |x-x| = 0\\$Hence by the principle of mathemtical induction our prposition has been proven.$\\\textbf{Case 2.2: } n = 2m + 1, \; m \in \mathbb{N}\\$Proposition: $ f_n(x) = -x\\$Proof: $ f_{2k+1}(x) = |x+f_{2k}(x)| = |x+0| = |x| = -x\\\\ \therefore \mathcal{I}_n = \int_{-1}^{1}f_n(x)\,dx = \int_{-1}^{0} f_n(x)\,dx + \int_{0}^{1} f_n(x)\,dx = \frac{n}{2} $ in the case that $n$ is even, or $ \frac{n+1}{2} $ in the case that n is odd.$$

Sorry my formatting isn't the best

$Correct!$

$\textbf{NEW QUESTION}$

$Find $\int \frac{9 +6\sqrt{x} + x}{4\sqrt{x} + x}\text{ d}x$.$

porcupinetree · Nov 25, 2015

Re: MX2 2016 Integration Marathon

InteGrand said:
$Correct!$

$\textbf{NEW QUESTION}$

$Find $\int \frac{9 +6\sqrt{x} + x}{4\sqrt{x} + x}\text{ d}x$.$

I think I have it, let me write it up

porcupinetree · Nov 25, 2015

Re: MX2 2016 Integration Marathon

porcupinetree said:
I think I have it, let me write it up

dan964 · Nov 25, 2015

Re: MX2 2016 Integration Marathon

Ekman said:
$\frac{d}{dx} \left( \frac{e^x \cdot \sin{x}}{1-\cos{x}} \right) = \frac{e^x(\sin{x} - 1)}{1-\cos{x}}$

$\therefore \int \left (\frac{1 - \sin x}{1 - \cos x} \right ) e^x \, dx = - \left( \frac{e^x \cdot \sin{x}}{1-\cos{x}} \right) + c$

(This question reminds me of the last question of the 2014 hsc, where doing IBP, via inspection can solve it in a couple of steps)

what is IBP, is it integration by parts?

HSC 2016 MX2 Integration Marathon (archive) (4 Viewers)

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