# Thread: HSC 2018-2019 MX2 Integration Marathon

1. ## HSC 2018-2019 MX2 Integration Marathon

Integration marathon for MX2 2018
Post integration questions within scope of MX2

First question: (from pikachu975)
Integrate 1/[(sinx)sqrt(sin2x)]

2. ## Re: HSC 2018 MX2 Integration Marathon

Originally Posted by pikachu975
$\int \frac{\text{d}x}{\sin{x}\sqrt{\sin(2x)}}$
Hint: expand and substitute

3. ## Re: HSC 2018 MX2 Integration Marathon

Hint: expand and substitute
Further hint: take something out of the root and simplify, then substitute

4. ## Re: HSC 2018 MX2 Integration Marathon

Wait the sinx is on the denominator. It's basically the BOS trial question

5. ## Re: HSC 2018 MX2 Integration Marathon

Originally Posted by pikachu975
Wait the sinx is on the denominator. It's basically the BOS trial question
Yup. FYI the mistranscribed integral is equivalent to 1/root(2) integral of root(tanx) which has been calculated to exhaustion.

6. ## Re: HSC 2018 MX2 Integration Marathon

Originally Posted by Kingom
Yup. FYI the mistranscribed integral is equivalent to 1/√2 integral of √tanx which has been calculated to exhaustion.
$\int \sqrt{\tan{x}} \, \text{d}x \equiv \frac{1}{2} \int (\sqrt{\tan{x}}-\sqrt{\cot{x}}) + (\sqrt{\tan{x}}+\sqrt{\cot{x}}) \, \text{d}x$

The rest is trivial.

7. ## Re: HSC 2018 MX2 Integration Marathon

$\noindent Evaluate \int_{-1}^1 \frac{e^{2x} + 1 - (x + 1)(e^x + e^{-x})}{x(e^x -1)} \, dx.$

8. ## Re: HSC 2018 MX2 Integration Marathon

For $x\geqslant 0$, define $f(x)=\int_{0}^{\pi}t\,sin^{x}t\,dt$.
(a)Show that f(x) is decreasing. Write down the maximum value of f(x).
(b)Find $\int \ln\,f(x) - \ln\,f(x+2)\,dx$.

9. ## Re: HSC 2018 MX2 Integration Marathon

$\noindent Evaluate \int_{-1}^1 \frac{e^{2x} + 1 - (x + 1)(e^x + e^{-x})}{x(e^x -1)} \, dx.$
\begin{align*}I&= \int_{-1}^1 \frac{e^{2x}+1 - (x+1)(e^x+e^{-x})}{x(e^x-1)}dx\\ &= \int_{-1}^1 \frac{(e^x-x-1)(e^x+e^{-x})}{x(e^x-1)}dx\\ &= -\int_{1}^{-1}\frac{(e^{-u}+u-1)(e^u+e^{-u})}{-u(e^{-u}-1)}du\\ &= \int_{-1}^1 \frac{(1+ue^u-e^u)(e^u+e^{-u})}{u(e^u-1)}du\end{align*}

\begin{align*}\therefore 2I &= \int_{-1}^1 \frac{e^x+e^{-x}}{x(e^x-1)}[e^x-x-1+1+xe^x-e^x]dx\\ &= \int_{-1}^1 \frac{e^x+e^{-x}}{xe^x-1}[xe^x-1]dx\\ &= \int_{-1}^1 (e^x+e^{-x})dx\\ \implies I&= e-\frac1e\end{align*}

10. ## Re: HSC 2018 MX2 Integration Marathon

Originally Posted by stupid_girl
For $x\geqslant 0$, define $f(x)=\int_{0}^{\pi}t\,sin^{x}t\,dt$.
(a)Show that f(x) is decreasing. Write down the maximum value of f(x).
(b)Find $\int \ln\,f(x) - \ln\,f(x+2)\,dx$.
$\text{Suppose that }0\le x \le y$

$\\\text{Since }t\in [0,\pi] \implies \sin t \in [0,1]\\ x\le y \implies \sin^x t \ge \sin^y t$

$\\\text{Hence for }t\in [0,\pi]\\ t\sin^x t \ge t \sin^y t \implies \int_0^\pi t\sin^x t\,dt \ge \int_0^\pi t\sin^y t\,dt\implies f(x)\ge f(y)\\ \text{Hence }f\text{ is decreasing.}$

$\text{So the maximum value is }f(0) = \frac{\pi^2}{2}$
_________________________________________

\begin{align*}f(x+2) &= \int_0^\pi t \sin^{x+2} t\,dt \\ &= \int_0^\pi (t \sin t) \sin^{x+1} t\,dt \\ &= \left[(\sin t - t\cos t)\sin^{x+1}t\right]_0^\pi - \int_0^\pi (\sin t - t \cos t)(x+1)\cos t\sin^{x}t\,dt\\ &= -(x+1)\int_0^\pi \sin^{x+1}t\cos t\,dt + (x+1)\int_0^\pi t\cos^2 t \sin^{x}t\,dt\\ &= (x+1)\int_0^\pi t\sin^{x}t\,dt -(x+1)\int_0^\pi t\sin^{x+2} t\,dt\\ &= (x+1)f(x) - (x+1)f(x+2)\\ \implies \frac{f(x)}{f(x+2)}&= \frac{x+2}{x+1}\end{align*}

$\therefore \int \ln f(x) - \ln f(x+2)\,dx = \int \ln (x+2) - \ln (x+1)\, dx\\ \text{The remainder of the question is trivial and is left as an exercise to the reader.}$

11. ## Re: HSC 2018 MX2 Integration Marathon

$\int_0^\infty \frac{dx}{\left(x+\sqrt{1+x^2}\right)^n}$

12. ## Re: HSC 2018 MX2 Integration Marathon

Originally Posted by leehuan
$\int_0^\infty \frac{dx}{\left(x+\sqrt{1+x^2}\right)^n}$
same

13. ## Re: HSC 2018 MX2 Integration Marathon

Originally Posted by leehuan
$\int_0^\infty \frac{dx}{\left(x+\sqrt{1+x^2}\right)^n}$
$\noindent Note, the (improper) integral only converges for n > 1.$

$\noindent The simplest way to solve this integral is to use a hyperbolic substitution of x = \sinh t, though as hyperbolic functions are not considered part of the MX2 syllabus I will make do with using its definition in terms of exponential functions.$

$\noindent Let x = \frac{e^t - e^{-t}}{2}, dx = \frac{e^t + e^{-t}}{2} \, dt with the limits of integration remaining unchanged.$

$\noindent So for the term 1 + x^2, in terms of t we have$

$1 + x^2 = \left (\frac{e^t + e^{-t}}{2} \right )^2,$

$\noindent so the term in the denominator becomes$

$x + \sqrt{1 + x^2} = e^t.$

$\noindent Thus$

\begin{align*}\int^\infty_0 \frac{dx}{(x + \sqrt{1 + x^2})^n} &= \frac{1}{2} \int^\infty_0 \frac{e^t + e^{-t}}{e^{nt}} \, dt\\ &= \frac{1}{2} \int^\infty_0 \left (e^{(1 - n)t} + e^{-(1 + n)t} \right ) \, dt\\ &= \frac{1}{2} \left [\frac{1}{1 - n} e^{(1 - n)t} - \frac{1}{1 + n} e^{-(1 + n)t} \right ]^\infty_0.\end{align*}

$\noindent Now as n > 1, 1 - n < 0 so e^{(1 - n)t} \to 0 as t \to \infty. Also, as n > 1, n + 1 > 0 so e^{-(1 + n)t} \to 0 as t \to \infty. Thus$

$\int^\infty_0 \frac{dx}{(x + \sqrt{1 + x^2})^n} = -\frac{1}{2} \left [\frac{1}{1 - n} - \frac{1}{1 + n} \right ] = \frac{n}{n^2 - 1}.$

14. ## Re: HSC 2018 MX2 Integration Marathon

$\noindent Now, so as not to scare off the children too soon, how able something a little easier?$

$\noindent Find, without using a trigonometric substitution, \int \frac{dx}{x^2 \sqrt{x^2 + 1}}.$

15. ## Re: HSC 2018 MX2 Integration Marathon

$\text{For }a>0,\text{ find }\int_0^{\infty} \frac{dx}{(1+x^a)(1+x^2)}$

Also spoiler for the one above:
 Spoiler (rollover to view): Do the same substitution as the one above it

16. ## Re: HSC 2018 MX2 Integration Marathon

Originally Posted by leehuan
$\text{For }a>0,\text{ find }\int_0^{\infty} \frac{dx}{(1+x^a)(1+x^2)}$

Also spoiler for the one above:
 Spoiler (rollover to view): Do the same substitution as the one above it
I don't think

$u^2 = 1+\frac{1}{x^2}$

is the right substitution for that.

17. ## Re: HSC 2018 MX2 Integration Marathon

This one is fairly straight-forward.
$\int (sec x-tan x)^{\frac{2018}{2017}}d(tan x)$

18. ## Re: HSC 2018 MX2 Integration Marathon

A nice one
$\noindent Evaluate \int_{0}^{1}x\sqrt{x\sqrt[3]{x\sqrt[4]{x\sqrt[5]{x...}}}} dx$

19. ## Re: HSC 2018 MX2 Integration Marathon

Originally Posted by jathu123
A nice one
$\noindent Evaluate \int_{0}^{1}x\sqrt{x\sqrt[3]{x\sqrt[4]{x\sqrt[5]{x...}}}} dx$
sik MIT steal

20. ## Re: HSC 2018 MX2 Integration Marathon

sik MIT steal
That being said is there a way of doing it using only 4U methods

21. ## Re: HSC 2018 MX2 Integration Marathon

Originally Posted by leehuan
That being said is there a way of doing it using only 4U methods
not without assuming the standard series expansion of e

22. ## Re: HSC 2018 MX2 Integration Marathon

not without assuming the standard series expansion of e
Which isn't even 4U.

Savage,

23. ## Re: HSC 2018 MX2 Integration Marathon

Originally Posted by leehuan
Which isn't even 4U.

Savage,
We can prove series expansion of $e$ using 4U maths.

24. ## Re: HSC 2018 MX2 Integration Marathon

Originally Posted by sharky564
We can prove series expansion of $e$ using 4U maths.
using a highly unmotivated proof, sure.... xd

25. ## Re: HSC 2018 MX2 Integration Marathon

but in the hsc such a question would not be likely to appear, all that work just for a simple integral....

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