HSC 2016 MX2 Integration Marathon (archive) (3 Viewers)

leehuan · Nov 6, 2015

Re: MX2 2016 Integration Marathon

KingOfActing said:
NEXT QUESTION:
$\int{xe^{x}\sin{x}dx}$

You question skippers

So let's just clarify briefly that for this one, find the integral of e^(x)sin(x) and e^(x)cos(x) first using a 'loopy' integration by parts. Then, we use integration by parts on the given integral, treating x as the expression to be differentiated, then sub in our results immediately earlier.

leehuan · Nov 7, 2015

Re: MX2 2016 Integration Marathon

$\\ $Find a reduction formula for the integral ${ I }_{ n }=\int { \arcsin ^{ n }{ x } dx }$

Paradoxica · Nov 7, 2015

Re: MX2 2016 Integration Marathon

leehuan said:
You question skippers

So let's just clarify briefly that for this one, find the integral of e^(x)sin(x) and e^(x)cos(x) first using a 'loopy' integration by parts. Then, we use integration by parts on the given integral, treating x as the expression to be differentiated, then sub in our results immediately earlier.

Personally, I would differentiate those instead of integrating them. Then add/subtract them together to yield the desired integral. Such is the elegance of periodic functions.

glittergal96 · Nov 7, 2015

Re: MX2 2016 Integration Marathon

Paradoxica said:
Such is the elegance of periodic functions.

That example doesn't really have anything to do with periodicity.

It so happens that the functions sin(x) and cos(x) are periodic, but this property is pretty irrelevant to that integration.

Paradoxica · Nov 7, 2015

Re: MX2 2016 Integration Marathon

glittergal96 said:
That example doesn't really have anything to do with periodicity.

It so happens that the functions sin(x) and cos(x) are periodic, but this property is pretty irrelevant to that integration.

Such is the elegance of differentially periodic functions.

omegadot · Nov 8, 2015

Re: MX2 2016 Integration Marathon

leehuan said:
$\\ $Find a reduction formula for the integral ${ I }_{ n }=\int { \arcsin ^{ n }{ x } dx }$

$$IBP gives$

$I_n = x \arcsin^n x - n \int \frac{x}{\sqrt{1 - x^2}} \arcsin^{n - 1} x \, dx$

$By noting $\int \frac{x}{\sqrt{1 - x^2}} dx = -\sqrt{1 - x^2} + \cal{C}$ (use the substitution $x = \sin \theta$ to show this)$

$IBP again yields$

$I_n = x \arcsin^n x + n\sqrt{1 - x^2}\, \arcsin^{n - 1} x - n(n - 1) I_{n - 2}, \quad n \geqslant 2.$

$$Next question: Evaluate $\int^{2\pi}_0 \left |\sin x - \cos x \right | \, dx.$

leehuan · Nov 8, 2015

Re: MX2 2016 Integration Marathon

Very briefly, don't even need the substitution x=sin(theta) to verify your result. You can just let u=1-x^2.

porcupinetree · Nov 8, 2015

Re: MX2 2016 Integration Marathon

If any 2016'ers are brave enough...

$\int \sqrt{tanx}\ dx$

leehuan · Nov 8, 2015

Re: MX2 2016 Integration Marathon

porcupinetree said:
If any 2016'ers are brave enough...

$\int \sqrt{tanx}\ dx$

Oh come on. So early into the year and you're throwing THAT one at them

InteGrand · Nov 8, 2015

Re: MX2 2016 Integration Marathon

leehuan said:
Oh come on. So early into the year and you're throwing THAT one at them

A devilishly tedious integral.

omegadot · Nov 8, 2015

Re: MX2 2016 Integration Marathon

leehuan said:
Very briefly, don't even need the substitution x=sin(theta) to verify your result. You can just let u=1-x^2.

Oh yes. Silly me!

Paradoxica · Nov 8, 2015

Re: MX2 2016 Integration Marathon

porcupinetree said:
If any 2016'ers are brave enough...

$\int \sqrt{\tan{x}}\ dx$

$$Hint: Let porcupine$^2 = 2\tan{x}$

$$Additional tediosity: $\sqrt[3]{\tan{x}}$

kawaiipotato · Nov 9, 2015

Re: MX2 2016 Integration Marathon

$\int \frac{ \sin x}{ \sin x + \cos x} dx$

leehuan · Nov 9, 2015

Re: MX2 2016 Integration Marathon

$\\ I=\int { \frac { \sin { x } }{ \sin { x } +\cos { x } } dx } \\ I=-\int { \left( \frac { \cos { x } -\sin { x } }{ \sin { x } +\cos { x } } -\frac { \cos { x } }{ \sin { x } +\cos { x } } \right) dx } \\ I=-\int { \frac { \cos { x } -\sin { x } }{ \sin { x } +\cos { x } } dx } +\int { \frac { \cos { x } }{ \sin { x } +\cos { x } } dx } \\ I=-\ln { \left| \sin { x } +\cos { x } \right| } +\int { \frac { \cos { x } }{ \sin { x } +\cos { x } } dx } \\ \therefore 2I=-\ln { \left| \sin { x } +\cos { x } \right| } +\int { \frac { \sin { x } +\cos { x } }{ \sin { x } +\cos { x } } dx } \\ I=\frac { 1 }{ 2 } \left( x-\ln { \left| \sin { x } +\cos { x } \right| } \right)$

(5th line achieved by adding lines 1 and 4)

Paradoxica · Nov 10, 2015

Re: MX2 2016 Integration Marathon

$$Evaluate $\int_0^{\infty} \frac{\tan^{-1}{x}}{x(x^2+1)^5}dx$

$$using the substitution $ y = \tan^{-1}{x}$

leehuan · Nov 10, 2015

Re: MX2 2016 Integration Marathon

NVM found the mistake. Lol, sec^2 lee.

InteGrand · Nov 10, 2015

Re: MX2 2016 Integration Marathon

leehuan said:
NVM found the mistake. Lol, sec^2 lee.

$Note that the process would be to take the limit as $u\to 0^+$, so $u\ln \left|\sin u\right|\to 0$ when evaluating the lower limit.$

leehuan · Nov 10, 2015

Re: MX2 2016 Integration Marathon

InteGrand said:
$Note that the process would be to take the limit as $u\to 0^+$, so $u\ln \left|\sin u\right|\to 0$ when evaluating the lower limit.$

Apart from L'Hopital's rule, is there any direct means of evaluating that limit? I did consider the limit but wasn't sure what to do within MX2

InteGrand · Nov 10, 2015

Re: MX2 2016 Integration Marathon

leehuan said:
Apart from L'Hopital's rule, is there any direct means of evaluating that limit? I did consider the limit but wasn't sure what to do within MX2

I doubt this integral would be asked in HSC MX2 these days.

leehuan · Nov 10, 2015

Re: MX2 2016 Integration Marathon

InteGrand said:
I doubt this integral would be asked in HSC MX2 these days.

I suppose so. Paradoxica loves putting in extracurricular integrals here.

HSC 2016 MX2 Integration Marathon (archive) (3 Viewers)

Well-Known Member

Well-Known Member

-insert title here-

Active Member

-insert title here-

Active Member

Well-Known Member

not actually a porcupine

Well-Known Member

Well-Known Member

Active Member

-insert title here-

Well-Known Member

Well-Known Member

-insert title here-

Well-Known Member

Well-Known Member

Well-Known Member

Well-Known Member

Well-Known Member

Users Who Are Viewing This Thread (Users: 0, Guests: 3)