HSC 2017 MX2 Integration Marathon (archive) (5 Viewers)

InteGrand · Jan 4, 2017

Re: HSC 4U Integration Marathon 2017

InteGrand said:
$$\noindent Another method is as follows (we can replace $2016$ with any non-negative even integer $N$ and $\pi$ in the integrand with any positive real constant $a$, so we will do this).$

$\noindent Let $I:= \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \frac{\left(2+3a^{x}\right)\sin^{N}(x) +\left(4 +5a^x \right)\cos^{N}x}{\left(1+a^{x}\right) \left(\sin^{N}x + \cos^{N} x\right)}\, \mathrm{d}x$. Then we have$

$I = \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \frac{\left(2a^{x}+3\right)\sin^{N}(x) +\left(4a^{x} +5\right)\cos^{N}x}{\left(1+a^{x}\right) \left(\sin^{N}x + \cos^{N} x\right)}\, \mathrm{d}x. \quad (\star)$

$\noindent So addition yields:$

$\begin{align*}2I &= 5\times \frac{\pi}{2} + 9\times \frac{\pi}{2}\quad (\star \star) \\ &= 7\pi \\ \Rightarrow I &= \frac{7\pi}{2}.\end{align*}$

$\noindent Justification of steps is left as an exercise for the reader, but some hints are given in the next post.$

$\noindent Equation $(\star)$ follows from using $\int_{-b}^{b}f(x)\,\mathrm{d}x = \int_{-b}^{b}f(-x)\,\mathrm{d}x$.$

$\noindent To help show $(\star \star)$, recall or prove that $\int_{0}^{\frac{\pi}{2}} \frac{1}{1+\tan^{\alpha}x}\,\mathrm{d}x = \frac{\pi}{4}$ for any real constant $\alpha$.$

calamebe · Jan 4, 2017

Re: HSC 4U Integration Marathon 2017

InteGrand said:
$$\noindent Another method is as follows (we can replace $2016$ with any non-negative even integer $N$ and $\pi$ in the integrand with any non-negative real constant $a$, so we will do this).$

$\noindent Let $I:= \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \frac{\left(2+3a^{x}\right)\sin^{N}(x) +\left(4 +5a^x \right)\cos^{N}x}{\left(1+a^{x}\right) \left(\sin^{N}x + \cos^{N} x\right)}\, \mathrm{d}x$. Then we have$

$I = \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \frac{\left(2a^{x}+3\right)\sin^{N}(x) +\left(4a^{x} +5\right)\cos^{N}x}{\left(1+a^{x}\right) \left(\sin^{N}x + \cos^{N} x\right)}\, \mathrm{d}x. \quad (\star)$

$\noindent So addition yields:$

$\begin{align*}2I &= 5\times \frac{\pi}{2} + 9\times \frac{\pi}{2}\quad (\star \star) \\ &= 7\pi \\ \Rightarrow I &= \frac{7\pi}{2}.\end{align*}$

$\noindent Justification of steps is left as an exercise for the reader, but some hints are given in the next post.$

Ah yes that makes sense, and so much simpler than my proof. Thank you!

stupid_girl · Jan 23, 2017

Re: HSC 4U Integration Marathon 2017

This one is a bit tedious.

$\int_{0}^{\frac{\pi^2}{16}} tan\sqrt{x}+\frac{2}{1+tan\sqrt{x}}$

calamebe · Jan 23, 2017

Re: HSC 4U Integration Marathon 2017

stupid_girl said:
This one is a bit tedious.

$\int_{0}^{\frac{\pi^2}{16}} tan\sqrt{x}+\frac{2}{1+tan\sqrt{x}}$

http://imgur.com/a/X7CkW

EDIT: Yeah I have no idea how to make the LaTeX valid here imgur will have to do haha. Also I'm still learning so if I can improve anywhere just tell me what to do.

$\begin{align*} \int_{0}^{\frac{\pi^2}{16}} \left(tan\sqrt{x}+\frac{2}{1+tan\sqrt{x}}\right)dx \end{align*}$

$Let $ u^2 = x \Rightarrow dx=2udu$$

$\begin{align*} \therefore \int_{0}^{\frac{\pi^2}{16}} tan\sqrt{x}+\frac{2}{1+tan\sqrt{x}}dx = \int_0^{\frac{\pi}{4}}2u\left(tan(u)+\frac{2}{1+tanu}\right)du \end{align*}$

$\begin{align*} \int_{0}^{\frac{\pi}{4}}2utanudu&=-2uln(cosu)\Big|_0^\frac{\pi}{4} + 2\int_{0}^{\frac{\pi}{4}}lncosudu\\ &=\tfrac{\pi}{4}ln2 +2\int_{0}^\frac{\pi}{4}lncosudu \end{align*}$

$$\noindent \begin{align*} \int\frac{du}{1+tanu}&=\int\frac{cosudu}{cosu+sinu}\\&=\int\left(\frac{\tfrac{1}{2}cosu+\tfrac{1}{2}sinu}{sinu+cosu}+\frac{\tfrac{1}{2}cosu-\tfrac{1}{2}sinu}{sinu+cosu}\right)du &=\tfrac{1}{2}\left(x+ln(cosu+sinu)\right)+C \end{align*} \begin{align*} \therefore \int_0^\frac{\pi}{4}\frac{4udu}{1+tanu} &= 2u(u+ln(cosu+sinu))\Big|_0^\frac{\pi}{4}-\int_0^\frac{\pi}{4}2udu -2\int_0^\frac{\pi}{4}ln(cosu+sinu)du\\ &=\frac{\pi}{8}\left(\pi+2ln2\right)-\frac{\pi^2}{16}-2\int_0^\frac{\pi}{4}ln(cos(\frac{\pi}{4}-u)+sin(\frac{\pi}{4}-u))du\\ &=\frac{\pi}{16}\left(\pi+4ln2\right) -2\int_{0}^\frac{\pi}{4}ln\sqrt{2}cosudu\\ &=\frac{\pi}{16}\left(\pi+4ln2\right) -2\int_{0}^\frac{\pi}{4}lncosudu-2\int_{0}^\frac{\pi}{4}ln\sqrt{2}du\\ &=\frac{\pi}{16}\left(\pi\right) -2\int_{0}^\frac{\pi}{4}lncosudu \end{align*}$
$\begin{align*} \therefore \int_{0}^{\frac{\pi^2}{16}} \left(tan\sqrt{x}+\frac{2}{1+tan\sqrt{x}}\right)dx = \frac{\pi}{16}(\pi+4ln(2)) \end{align*}$

InteGrand · Jan 23, 2017

Re: HSC 4U Integration Marathon 2017

calamebe said:
http://imgur.com/a/X7CkW

EDIT: Yeah I have no idea how to make the LaTeX valid here imgur will have to do haha. Also I'm still learning so if I can improve anywhere just tell me what to do.

Still learning what? LaTeX, or Integration?

calamebe · Jan 23, 2017

Re: HSC 4U Integration Marathon 2017

InteGrand said:
Still learning what? LaTeX, or Integration?

LaTeX

EDIT: Though I guess I can always get better at integration and it doesn't stop at high school

InteGrand · Jan 23, 2017

Re: HSC 4U Integration Marathon 2017

calamebe said:
LaTeX

I guess you can post your LaTeX attempts (or do it in the LaTeX Practice Thread) and people may give LaTeX tips then (since then we can see what's wrong with the code, if anything).

wu345 · Feb 7, 2017

Re: HSC 2017 MX2 Integration Marathon

By considering
$\int_{0}^{\infty} f\left((x-x^{-1})^{2}\right) dx$

evaluate:
$\int_{0}^{\infty} \frac{x^{2n-2}}{(x^4-x^{2}+1)^{n}} dx$

Paradoxica · Feb 15, 2017

Re: HSC 4U Integration Marathon 2017

stupid_girl said:
This one is a bit tedious.

$\int_{0}^{\frac{\pi^2}{16}} tan\sqrt{x}+\frac{2}{1+tan\sqrt{x}}$

I disagree. With the initial substitution $u^2=x$ , the standard reflection formula may be applied and we obtain the following equivalence after simplification:

$I=2 \int_0^\frac{\pi}{4} u \left(\tan{u}+\frac{2}{1+\tan{u}}\right) \text{d}u=2\int_0^\frac{\pi}{4} \left(\frac{\pi}{4}-u\right) \left(\tan{u}+\frac{2}{1+\tan{u}}\right)\text{d}u$

$I=\frac{\pi}{4}\int_0^\frac{\pi}{4} \left(\tan{u}+\frac{2}{1+\tan{u}}\right)\text{d}u$

Split and reflect the integral again, but carefully, as follows:

$\tan{u}+\frac{2}{1+\tan{u}} = \tan{u} + \tan{\left(\frac{\pi}{4}-u \right)} + 1$

The value of the integral now simplifies into:

$I = \frac{\pi}{2}\int_0^\frac{\pi}{4} \tan{u} \, \text{d}u + \frac{\pi^2}{16}=\frac{\pi\log{2}}{2} + \frac{\pi^2}{16}$

Paradoxica · Feb 15, 2017

Re: HSC 2017 MX2 Integration Marathon

wu345 said:
By considering
$\int_{0}^{\infty} f\left((x-x^{-1})^{2}\right) dx$

evaluate:
$\int_{0}^{\infty} \frac{x^{2n-2}}{(x^4-x^{2}+1)^{n}} dx$

Using the reflection across the real line with u+x=0, and Glasser's Master Theorem (ceebs showing the intended way but KingOfActing can do that), the integral simplifies completely into:

$\int_0^\infty \frac{\text{d}x}{(x^2+1)^n}$

This can easily be evaluated in the general form using Differentiation Under The Integral Sign.

BenHowe · Feb 22, 2017

Re: HSC 2017 MX2 Integration Marathon

$\text{1.}\int\sqrt{-4x^{2}+3x+6}dx\\\text{2.}\int\frac{e^{x}}{1+e^{x}}dx\\\text{3.}\int\frac{xe^{x}}{1+e^{x}}dx\\\text{4.This was actually the last q in a hsc paper somewhere and involves a weird trick}\int\frac{lnx}{(1+lnx)^{2}}dx$

Drongoski · Feb 22, 2017

Re: HSC 2017 MX2 Integration Marathon

BenHowe said:
$\text{1.}\int\sqrt{-4x^{2}+3x+6}dx\\\text{2.}\int\frac{e^{x}}{1+e^{x}}dx\\\text{3.}\int\frac{xe^{x}}{1+e^{x}}dx\\\text{4.This was actually the last q in a hsc paper somewhere and involves a weird trick}\int\frac{lnx}{(1+lnx)^{2}}dx$

No 2 is easy:

$= \int \frac{d(e^x +1)}{e^x + 1} = \int \frac {d (apple)}{apple} = ln(apple) + C= ln(e^x + 1) + C$

si2136 · Feb 22, 2017

Re: HSC 2017 MX2 Integration Marathon

Can't you do two by inspection?

Green Yoda · Feb 22, 2017

Re: HSC 2017 MX2 Integration Marathon

Reckon q2 is 2u?

Drongoski · Feb 22, 2017

Re: HSC 2017 MX2 Integration Marathon

Rathin said:
Reckon q2 is 2u?

Perhaps. That's why I can do this one.

Drongoski · Feb 22, 2017

Re: HSC 2017 MX2 Integration Marathon

si2136 said:
Can't you do two by inspection?

Yes. I did it by inspection but had to show the steps.

pikachu975 · Feb 22, 2017

Re: HSC 2017 MX2 Integration Marathon

BenHowe said:
$\text{1.}\int\sqrt{-4x^{2}+3x+6}dx\\\text{2.}\int\frac{e^{x}}{1+e^{x}}dx\\\text{3.}\int\frac{xe^{x}}{1+e^{x}}dx\\\text{4.This was actually the last q in a hsc paper somewhere and involves a weird trick}\int\frac{lnx}{(1+lnx)^{2}}dx$

http://m.imgur.com/j3kDVl4

Q1 answer, it's messy so if you can't understand it or read it then tell me. I don't know if there's a faster method but if there is please tell me.

BenHowe · Feb 22, 2017

Re: HSC 2017 MX2 Integration Marathon

pikachu975 said:
http://m.imgur.com/j3kDVl4

Q1 answer, it's messy so if you can't understand it or read it then tell me. I don't know if there's a faster method but if there is please tell me.

Correct! One of those ones where you have to complete the square then do the appropriate trig sub and fiddle with a right angled triangle...

stupid_girl · Feb 23, 2017

Re: HSC 2017 MX2 Integration Marathon

Are you sure Q3 has closed form solution?

BenHowe · Feb 23, 2017

Re: HSC 2017 MX2 Integration Marathon

I don't know what closed form is but you just need to be creative

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