HSC 2015 MX2 Integration Marathon (archive) (1 Viewer)

glittergal96 · Jul 14, 2015

Re: MX2 2015 Integration Marathon

InteGrand said:
Probably a typo or something, but we should have +(n – 2) there in the numerator.

Yep, yep, +. Cheers

.

leehuan · Jul 15, 2015

Re: MX2 2015 Integration Marathon

$\\ $Let ${ I }_{ n }=\int _{ 1 }^{ 2 }{ { \left( 1-\frac { 1 }{ x } \right) }^{ n }dx } \\ $Show that $\frac { 1 }{ n+1 } { I }_{ n+1 }=\frac { 1 }{ n } { I }_{ n }-\frac { 1 }{ n\left( n+1 \right) { 2 }^{ n } }$
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Lol, ok I'll do this later myself.

leehuan said:
$\int { \sin ^{ -1 }{ \sqrt { x } } \cos ^{ -1 }{ \sqrt { x } } dx }$

Drsoccerball · Jul 15, 2015

Re: MX2 2015 Integration Marathon

leehuan said:
$\\ $Let ${ I }_{ n }=\int _{ 1 }^{ 2 }{ { \left( 1-\frac { 1 }{ x } \right) }^{ n }dx } \\ $Show that $\frac { 1 }{ n+1 } { I }_{ n+1 }=\frac { 1 }{ n } { I }_{ n }-\frac { 1 }{ n\left( n+1 \right) { 2 }^{ n } }$
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Lol, ok I'll do this later myself.

I posted a similar question where $\sqrt{x} \rightarrow x$

leehuan · Jul 15, 2015

Re: MX2 2015 Integration Marathon

Square roots make the questions more interesting.

$\\ I=\int { \arcsin { \sqrt { x } } \arccos { \sqrt { x } } dx } \\ =\frac { \pi }{ 2 } \int { \arcsin { \sqrt { x } } dx } -\int { \arcsin ^{ 2 }{ \sqrt { x } } dx } \\ Let\quad u=\arcsin { \sqrt { x } } \Rightarrow x=\sin ^{ 2 }{ u } \Rightarrow dx=2\sin { u } \cos { u } du=\sin { 2u } du\\ I=\frac { \pi }{ 2 } \int { u\sin { 2u } du } -\int { { u }^{ 2 }\sin { 2u } du } \\ Three\quad applications\quad of\quad I.B.P.:$

$\\ I=\frac { \pi }{ 8 } \left( \sin { 2u } -2u\cos { 2u } \right) -\frac { 1 }{ 2 } u\sin { 2u } +\frac { 1 }{ 4 } \left( 2{ u }^{ 2 }-1 \right) \cos { \left( 2u \right) } +C\\ \cos { u } =\sqrt { 1-{ x }^{ 2 } } ,\sin { u } =\sqrt { x } \\ \sin { 2u } =2\sin { u } \cos { u } =2\sqrt { x-{ x }^{ 3 } } \\ \cos { 2u } =1-2\sin ^{ 2 }{ u } =1-2x\\ I=\frac { \pi }{ 8 } \left( 2\sqrt { x-{ x }^{ 3 } } -2\left( 1-2x \right) \arcsin { \sqrt { x } } \right) -\frac { 1 }{ 2 } \sqrt { x-{ x }^{ 3 } } \arcsin { \sqrt { x } } +\frac { 1 }{ 4 } \left( 2\arcsin ^{ 2 }{ \sqrt x } -1 \right) \left( 1-2x \right) +C$

Drsoccerball · Jul 15, 2015

Re: MX2 2015 Integration Marathon

leehuan said:
Square roots make the questions more interesting.

$\\ I=\int { \arcsin { \sqrt { x } } \arccos { \sqrt { x } } dx } \\ =\frac { \pi }{ 2 } \int { \arcsin { \sqrt { x } } dx } -\int { \arcsin ^{ 2 }{ \sqrt { x } } dx } \\ Let\quad u=\arcsin { \sqrt { x } } \Rightarrow x=\sin ^{ 2 }{ u } \Rightarrow dx=2\sin { u } \cos { u } du=\sin { 2u } du\\ I=\frac { \pi }{ 2 } \int { u\sin { 2u } du } -\int { { u }^{ 2 }\sin { 2u } du } \\ Three\quad applications\quad of\quad I.B.P.:$

$\\ I=\frac { \pi }{ 8 } \left( \sin { 2u } -2u\cos { 2u } \right) -\frac { 1 }{ 2 } u\sin { 2u } +\frac { 1 }{ 4 } \left( 2{ u }^{ 2 }-1 \right) \cos { \left( 2u \right) } +C\\ \cos { u } =\sqrt { 1-{ x }^{ 2 } } ,\sin { u } =\sqrt { x } \\ \sin { 2u } =2\sin { u } \cos { u } =2\sqrt { x-{ x }^{ 3 } } \\ \cos { 2u } =1-2\sin ^{ 2 }{ u } =1-2x\\ I=\frac { \pi }{ 8 } \left( 2\sqrt { x-{ x }^{ 3 } } -2\left( 1-2x \right) \arcsin { \sqrt { x } } \right) -\frac { 1 }{ 2 } \sqrt { x-{ x }^{ 3 } } \arcsin { \sqrt { x } } +\frac { 1 }{ 4 } \left( 2\arcsin ^{ 2 }{ \sqrt x } -1 \right) \left( 1-2x \right) +C$

Very ugly

Ekman · Jul 19, 2015

Re: MX2 2015 Integration Marathon

Here is a simple one to get the ball rolling again:

$\int_{0}^{\frac{\pi}{2}} \frac{1-\tan{x}}{1+\tan{x}} dx$

kawaiipotato · Jul 19, 2015

Re: MX2 2015 Integration Marathon

Ekman said:
Here is a simple one to get the ball rolling again:

$\int_{0}^{\frac{\pi}{2}} \frac{1-\tan{x}}{1+\tan{x}} dx$

$$ I $ = \int_0^{\frac{\pi}{2}} \frac{cosx - sinx}{cosx+sinx} dx$

$= [\ln(cosx+sinx)]_0^\frac{\pi}{2}$

$= 0$

-----------------

$New question$

$\int_0^{\frac{\pi}{2}} \frac{1}{1+tan^{\pi}(x)} dx$

Drsoccerball · Jul 19, 2015

Re: MX2 2015 Integration Marathon

kawaiipotato said:
$$ I $ = \int_0^{\frac{\pi}{2}} \frac{cosx - sinx}{cosx+sinx} dx$

$= [\ln(cosx+sinx)]_0^\frac{\pi}{2}$

$= 0$

-----------------

$New question$

$\int_0^{\frac{\pi}{2}} \frac{1}{1+tan^{\pi}(x)} dx$

$I=\int_0^{\frac{\pi}{2}}{\frac{1}{1+tan^{\pi}(x)}}dx= \int_0^{\frac{\pi}{2}}{\frac{1}{1+cot^{\pi}(x)}}$

$2I=\int_0^{\frac{\pi}{2}}1dx$

$I=\frac{\pi}{4}$

porcupinetree · Jul 19, 2015

Re: MX2 2015 Integration Marathon

Another fairly easy one:

$\int_{2}^{\infty } \frac{dx}{\sqrt{x-1} + \sqrt{(x-1)^{3}}}$

Drsoccerball · Jul 19, 2015

Re: MX2 2015 Integration Marathon

porcupinetree said:
Another fairly easy one:

$\int_{2}^{\infty } \frac{dx}{\sqrt{x-1} + \sqrt{(x-1)^{3}}}$

Similar to the one at the top of this page:
$$ let $ u^6=(x-1)$

$\int_1^{\infty}{\frac{6u^2}{1+u^6}}$

$2[tan^{-1}u^3]_1^{\infty}$

$\frac{\pi}{2}$

Drsoccerball · Jul 19, 2015

Re: MX2 2015 Integration Marathon

$\int_0^1{\frac{1}{1-log_2(1-x)}}$

kawaiipotato · Jul 19, 2015

Re: MX2 2015 Integration Marathon

Two more questions
$\int_0^1 2^{log_e(x)} dx$

$\int_0^1 \log^{log_ex}(2) dx$

rand_althor · Jul 19, 2015

Re: MX2 2015 Integration Marathon

kawaiipotato said:
$\int_0^1 2^{log_e(x)} dx$

$\begin{align*}\int^1_0 2^{\ln{x}}\, \mathrm{d}x&=\int^1_0 e^{\ln{2^{\ln x}}} \, \mathrm{d}x \\&= \int^1_0 e^{\ln x \ln 2} \, \mathrm{d}x \\&=\int^1_0 x^{\ln 2} \, \mathrm{d}x \\&=\left [\frac{x^{\ln(2) +1}}{\ln(2)+1} \right ]^1_0 \\&=\frac{1^{\ln(2) +1}}{\ln(2)+1}-\frac{0^{\ln(2) +1}}{\ln(2)+1} \\&=\frac{1}{\ln(2)+1}\end{align*}$

porcupinetree · Jul 19, 2015

Re: MX2 2015 Integration Marathon

kawaiipotato said:
$\int_0^1 2^{log_e(x)} dx$

Edit - whoops didn't see that the question had already been done

porcupinetree · Jul 20, 2015

Re: MX2 2015 Integration Marathon

Drsoccerball said:
$\int_0^1{\frac{1}{1-log_2(1-x)}}$

Can't seem to crack this one - I did f(x) = f(a-x) trick, but that didn't really lead anywhere

Drsoccerball · Jul 20, 2015

Re: MX2 2015 Integration Marathon

I got to:
$A \int_0^1{\frac{1}{A-lnx}} $ Where A $ = \frac{log2}{loge}$ and i got stuck can anyone evaluate

Ekman · Jul 20, 2015

Re: MX2 2015 Integration Marathon

Drsoccerball said:
I got to:
$A \int_0^1{\frac{1}{A-lnx}} $ Where A $ = \frac{log2}{loge}$ and i got stuck can anyone evaluate

Did you confirm that it was possible to solve because Wolfram seems to be using li, which isn't a HSC concept.

porcupinetree · Jul 20, 2015

Re: MX2 2015 Integration Marathon

Drsoccerball said:
I got to:
$A \int_0^1{\frac{1}{A-lnx}} $ Where A $ = \frac{log2}{loge}$ and i got stuck can anyone evaluate

Yeah I got that too and couldn't find a way to evaluate. Then I looked on wolfram alpha as Ekman did and noted that it used the logarithmic integral

kawaiipotato · Jul 21, 2015

Re: MX2 2015 Integration Marathon

leehuan said:
$\\ $Let ${ I }_{ n }=\int _{ 1 }^{ 2 }{ { \left( 1-\frac { 1 }{ x } \right) }^{ n }dx } \\ $Show that $\frac { 1 }{ n+1 } { I }_{ n+1 }=\frac { 1 }{ n } { I }_{ n }-\frac { 1 }{ n\left( n+1 \right) { 2 }^{ n } }$

$$ let $ u = (1-\frac{1}{x})^{n} ... dv = dx$

$du = \frac{n}{x^2}(1-\frac{1}{x})^{n-1} ... v = x$

$\therefore I_n = [x(1-\frac{1}{x})^{n}]_1^2 - n \int_1^2 \frac{1}{x}(1-\frac{1}{x})^{n-1} dx$

$I_n = 2^{1-n} - n \int_1^2 (1-\frac{1}{x})^{n-1} - (1-\frac{1}{x})^{n} dx$

$I_n = \frac{1}{2^{n-1}} - n [I_{n-1} - I_{n}]$

$sub n \rightarrow n+1$

$I_{n+1} = \frac{1}{2^n} - (n+1)I_{n} + (n+1)I_{n+1}$

$nI_{n+1} = \frac{1}{2^n} - (n+1)I_{n}$

$\frac{1}{n+1}I_{n+1} = \frac{1}{n(n+1)2^n} - \frac{1}{n}I_{n}$

No idea why the signs are reversed, I checked multiple times.

InteGrand · Jul 21, 2015

Re: MX2 2015 Integration Marathon

kawaiipotato said:
$$ let $ u = (1-\frac{1}{x})^{n} ... dv = dx$

$du = \frac{n}{x^2}(1-\frac{1}{x})^{n} ... v = x$

$\therefore I_n = [x(1-\frac{1}{x})^{n}]_1^2 - n \int_1^2 \frac{1}{x}(1-\frac{1}{x})^{n-1} dx$

$I_n = 2^{1-n} - n \int_1^2 (1-\frac{1}{x})^{n-1} - (1-\frac{1}{x})^{n} dx$

$I_n = \frac{1}{2^{n-1}} - n [I_{n-1} - I_{n}]$

$sub n \rightarrow n+1$

$I_{n+1} = \frac{1}{2^n} - (n+1)I_{n} + (n+1)I_{n+1}$

$nI_{n+1} = \frac{1}{2^n} - (n+1)I_{n}$

$\frac{1}{n+1}I_{n+1} = \frac{1}{n(n+1)2^n} - \frac{1}{n}I_{n}$

No idea why the signs are reversed, I checked multiple times.

$When differentiating $u$, you forgot to lower the power.$

$Should be $\text{d}u=\frac{n}{x^2}\left(1-\frac{1}{x} \right)^{n-1}\text{ d}x$.$

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