HSC 2015 MX2 Integration Marathon (archive) (5 Viewers)

juantheron · Aug 30, 2015

Re: MX2 2015 Integration Marathon

$\displaystyle \int_{0}^{\frac{\pi}{4}}\frac{\cot x}{\cot^2 x+\sqrt{\cot x}}dx$

Drsoccerball · Aug 30, 2015

Re: MX2 2015 Integration Marathon

juantheron said:
$\displaystyle \int_{0}^{\frac{\pi}{4}}\frac{\cot x}{\cot^2 x+\sqrt{\cot x}}dx$

So if you rearrange you get :
$\int_0^{\frac{\pi}{4}}{\frac{\sqrt{cotx}}{cot^{3/2}x+1}}dx$

Let u^2=cotx and after that your integral becomes:

$\frac{-1}{2} \int{\frac{u^4+1}{u^3+1}}$

$$ Use plus minus technique to obtain $ \int{u- \frac{u}{u^3+1}}$

$Then use partial fractions to evaluate the second part ceebs doing the rest$

lita1000 · Aug 30, 2015

Re: MX2 2015 Integration Marathon

Drsoccerball said:
So if you rearrange you get :
$\int_0^{\frac{\pi}{4}}{\frac{\sqrt{cotx}}{cot^{3/2}x+1}}dx$

Let u^2=cotx and after that your integral becomes:

$\frac{-1}{2} \int{\frac{u^4+1}{u^3+1}}$

$$ Use plus minus technique to obtain $ \int{u- \frac{u}{u^3+1}}$

$Then use partial fractions to evaluate the second part ceebs doing the rest$

Can you really make that substitution when clearly cot 0 is undefined?

Also, can you show ur working for how you went from the trig to the u bit? I don't think ur integrand is quite right....

InteGrand · Aug 30, 2015

Re: MX2 2015 Integration Marathon

lita1000 said:
Can you really make that substitution when clearly cot 0 is undefined?

Also, can you show ur working for how you went from the trig to the u bit? I don't think ur integrand is quite right....

$The integral will be improper, but I get this after the substitution:$

$$I=\int_1 ^\infty \frac{2u^2}{(u^3+1)(u^4 +1)}\text{ d}u$.$

lita1000 · Aug 30, 2015

Re: MX2 2015 Integration Marathon

InteGrand said:
$The integral will be improper, but I get this after the substitution:$

$$I=\int_1 ^\infty \frac{2u^2}{(u^3+1)(u^4 +1)}$.$

That's exactly what i got, except with a negative sign

leehuan · Aug 30, 2015

Re: MX2 2015 Integration Marathon

InteGrand said:
$The integral will be improper, but I get this after the substitution:$

$$I=\int_1 ^\infty \frac{2u^2}{(u^3+1)(u^4 +1)}\text{ d}u$.$

I had this as well. But the partial fraction decomposition for this is ridiculously tedious.

InteGrand · Aug 30, 2015

Re: MX2 2015 Integration Marathon

lita1000 said:
That's exactly what i got, except with a negative sign

$The negative sign should be used to obtain the limits the way I put them. Do you mean you got this?:$

$\int _{\infty} ^1 \frac{-2u^2}{(u^3 + 1)(u^4+1)}\text{ d}u$

lita1000 · Aug 30, 2015

Re: MX2 2015 Integration Marathon

whoops yea i forgot about ur bounds

leehuan · Aug 30, 2015

Re: MX2 2015 Integration Marathon

Drsoccerball said:
So if you rearrange you get :
$\int_0^{\frac{\pi}{4}}{\frac{\sqrt{cotx}}{cot^{3/2}x+1}}dx$

Let u^2=cotx and after that your integral becomes:

$\frac{-1}{2} \int{\frac{u^4+1}{u^3+1}}$

$$ Use plus minus technique to obtain $ \int{u- \frac{u}{u^3+1}}$

$Then use partial fractions to evaluate the second part ceebs doing the rest$

${ u }^{ 2 }=\cot { x } \Rightarrow 2udu=-\csc ^{ 2 }{ x } dx\Rightarrow dx=-\frac { 2udu }{ 1+{ u }^{ 4 } } (u>0)\\ \int _{ 0 }^{ \frac { \pi }{ 4 } }{ \frac { \cot { x } }{ \cot ^{ 2 }{ x } +\sqrt { \cot { x } } } dx } \\ =\int _{ 0 }^{ \frac { \pi }{ 4 } }{ \frac { \sqrt { \cot { x } } }{ \cot ^{ \frac { 3 }{ 2 } }{ x } +1 } dx } \\ =\int _{ \infty }^{ 1 }{ \frac { u }{ { u }^{ 3 }+1 } \cdot -\frac { 2udu }{ 1+{ u }^{ 4 } } }$

Hm..
--------
But yeah, unpleasant. http://www.wolframalpha.com/input/?i=integrate+2u^2/((u^3+1)(u^4+1))

Drsoccerball · Aug 30, 2015

Re: MX2 2015 Integration Marathon

leehuan said:
${ u }^{ 2 }=\cot { x } \Rightarrow 2udu=-\csc ^{ 2 }{ x } dx\Rightarrow dx=-\frac { 2udu }{ 1+{ u }^{ 4 } } (u>0)\\ \int _{ 0 }^{ \frac { \pi }{ 4 } }{ \frac { \cot { x } }{ \cot ^{ 2 }{ x } +\sqrt { \cot { x } } } dx } \\ =\int _{ 0 }^{ \frac { \pi }{ 4 } }{ \frac { \sqrt { \cot { x } } }{ \cot ^{ \frac { 3 }{ 2 } }{ x } +1 } dx } \\ =\int _{ \infty }^{ 1 }{ \frac { u }{ { u }^{ 3 }+1 } \cdot -\frac { 2udu }{ 1+{ u }^{ 4 } } }$

Hm..

The algebra mistakes are real

Ekman · Aug 30, 2015

Re: MX2 2015 Integration Marathon

Drsoccerball said:
The algebra mistakes are real

$$Im$($mistakes$) = 0$

Paradoxica · Aug 31, 2015

Re: MX2 2015 Integration Marathon

Evaluate
$\int_{-\infty}^{\infty}{\frac{\sin^2{ax}}{x^2}}dx$
" $a$ " is an arbitrary constant

Drsoccerball · Aug 31, 2015

Re: MX2 2015 Integration Marathon

Paradoxica said:
Evaluate
$\int_{-\infty}^{\infty}{\frac{\sin^2{ax}}{x^2}}dx$
" $a$ " is an arbitrary constant

This was beautiful

Drsoccerball · Aug 31, 2015

Re: MX2 2015 Integration Marathon

Paradoxica said:
Evaluate
$\int_{-\infty}^{\infty}{\frac{\sin^2{ax}}{x^2}}dx$
" $a$ " is an arbitrary constant

$I= \lim_{b \rightarrow \infty} \int_{-b}^b{\frac{sin^2{ax}}{x^2}}$

$I=\int{\frac{1-cos^2{ax}}{x^2}}$

$2I= \int{\frac{1-cos{2ax}}{x^2}}$

$I= \frac{-1}{2} \int{\frac{cos{2ax}}{x^2}} $ Integrated the first part $$

$let x= \frac{cos^{-1}u}{2a}$

$$ New I $= a^2 \int{\frac{1}{\sqrt{1-u^2}(cos^{-1}u)^2}$

$$ Let cos^{-1}{u} =w$

$$ New I=-a^2 \int{\frac{1}{w^2}}dw$

$$ Final value $ = \frac{a^2}{w} =\frac{a^2}{cos^{-1}u}=\frac{a}{2x}$

$I=0$ LOL

glittergal96 · Aug 31, 2015

Re: MX2 2015 Integration Marathon

Lol it's positive everywhere, its integral over the real line is definitely NOT 0.

Drsoccerball · Aug 31, 2015

Re: MX2 2015 Integration Marathon

glittergal96 said:
Lol it's positive everywhere, its integral over the real line is definitely NOT 0.

the answers $\pi a$ but i cant see my mistake?

InteGrand · Aug 31, 2015

Re: MX2 2015 Integration Marathon

Drsoccerball said:
$2I= \int{\frac{1-cos{2ax}}{x^2}}$

$I= \frac{-1}{2} \int{\frac{cos{2ax}}{x^2}} $ Integrated the first part $$

$How did you ``integrate the first part''? Note that $\int _0 ^{b} \frac{1}{x^2}\text{ d}x$ does not converge for any positive $b$. Also, the definition of the improper integral ($I=2\int _{0}^\infty \frac{\sin^2 ax}{x^2}\text{ d}x$) should be something like $I=\lim_{\xi\to 0^+}\lim_{b\to +\infty}2\int _{\xi}^b \frac{\sin^2 ax}{x^2} \text{ d}x$, if the limit exists (which it does it turns out).$

Drsoccerball · Aug 31, 2015

Re: MX2 2015 Integration Marathon

InteGrand said:
$How did you ``integrate the first part''? Note that $\int _0 ^{b} \frac{1}{x^2}\text{ d}x$ does not converge for any positive $b$. Also, the definition of the improper integral ($I=2\int _{0}^\infty \frac{\sin^2 ax}{x^2}\text{ d}x$) should be something like $I=\lim_{\xi\to 0^+}\lim_{b\to +\infty}2\int _{\xi}^b \frac{\sin^2 ax}{x^2} \text{ d}x$, if the limit exists (which it does it turns out).$

I did it from $-\infty to\infty$

InteGrand · Aug 31, 2015

Re: MX2 2015 Integration Marathon

Drsoccerball said:
I did it from $-\infty to\infty$

That doesn't converge either.

Drsoccerball · Aug 31, 2015

Re: MX2 2015 Integration Marathon

InteGrand said:
That doesn't converge either.

Doesnt it give you 0 ? And also wot u mean by converging and diverging etc..

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