HSC 2015 MX2 Integration Marathon (archive) (2 Viewers)

Drsoccerball · Jun 11, 2015

Re: MX2 2015 Integration Marathon

For : $\int_0^1{\frac{1-x^2}{(1+x^2)\sqrt{1+x^4}}}$

$$ Divide top and bottom by $ x^2 $ and let $ u=\frac{1}{x} +x$

Drsoccerball · Jun 11, 2015

Re: MX2 2015 Integration Marathon

leehuan said:
Stumped on Q3.

Drsoccerball said:
For : $\int_0^1{\frac{1-x^2}{(1+x^2)\sqrt{1+x^4}}}$

$$ Divide top and bottom by $ x^2 $ and let $ u=\frac{1}{x} +x$

.

leehuan · Jun 11, 2015

Re: MX2 2015 Integration Marathon

My apologies. Somehow I matched that with Q2.

Drsoccerball · Jun 12, 2015

Re: MX2 2015 Integration Marathon

$\int{x^{2x}(1+lnx)}dx$

Ekman · Jun 12, 2015

Re: MX2 2015 Integration Marathon

Drsoccerball said:
$\int{x^{2x}(1+lnx)}dx$

Reverse Chain rule:

$\int{x^{2x}(1+lnx)}dx = \frac{x^{2x}}{2} +c$

Drsoccerball · Jun 13, 2015

Re: MX2 2015 Integration Marathon

$\int_0^{\pi/2}{\frac{1}{sin^4x +cos^4x}$

kawaiipotato · Jun 13, 2015

Re: MX2 2015 Integration Marathon

Drsoccerball said:
$\int_0^{\pi/2}{\frac{1}{sin^4x +cos^4x}$

$\int_0^{\pi/2}{\frac{dx}{sin^4x +cos^4x}$

$\int_0^{\pi/2}{\frac{dx}{(sin^{2}x + cos^{2}x)^{2} - 2cos^{2}xsin^{2}x}$

$\int_0^{\pi/2}{\frac{dx}{1 - \frac{1}{2}sin^{2}2x}$

$Let t = tan2x$

$\therefore \frac{t^2}{1+t^{2}} = sin^{2}2x$

$\int_0^1 \frac{dt}{2+t^{2}}$

$= [tan^{-1}(\frac{t}{\sqrt{2}})]_0^1$

Still didn't get the correct answer. Can someone check my method?

porcupinetree · Jun 13, 2015

Re: MX2 2015 Integration Marathon

kawaiipotato said:
$\int_0^{\pi/2}{\frac{dx}{sin^4x +cos^4x}$

$\int_0^{\pi/2}{\frac{dx}{(sin^{2}x + cos^{2}x)^{2} - 2cos^{2}xsin^{2}x}$

$\int_0^{\pi/2}{\frac{dx}{1 - \frac{1}{2}sin^{2}2x}$

$Let t = tan2x$

$\therefore \frac{t^2}{1+t^{2}} = sin^{2}2x$

$\int_0^1 \frac{dt}{2+t^{2}}$

$= [tan^{-1}(\frac{t}{\sqrt{2}})]_0^1$

Still didn't get the correct answer. Can someone check my method?

I'm confused by this question; whenever I try and make a substitution, the boundaries become equal (usually both become 0 - this is what should have happened to you in your second last line)

Drsoccerball · Jun 13, 2015

Re: MX2 2015 Integration Marathon

I did partial fractions and i got the answer...

porcupinetree · Jun 13, 2015

Re: MX2 2015 Integration Marathon

Drsoccerball said:
I did partial fractions and i got the answer...

Could you possibly post your working?

kawaiipotato · Jun 13, 2015

Re: MX2 2015 Integration Marathon

porcupinetree said:
I'm confused by this question; whenever I try and make a substitution, the boundaries become equal (usually both become 0 - this is what should have happened to you in your second last line)

I don't see why it should be equal in my second last line, I used substituion of t = tan2x, where x = 0 and x = pi/2

Drsoccerball · Jun 13, 2015

Re: MX2 2015 Integration Marathon

porcupinetree said:
Could you possibly post your working?

$I=\int_0^{\frac{\pi}{2}}{\frac{dx}{sin^4x +cos^4x}} = \frac{1}{2} \int_0^{\pi/2}{\frac{1}{sin^2x +cos^2x +\sqrt{2}sinxcosx}+ \frac{1}{sin^2x +cos^2x -\sqrt{2} cosxsinx}}$

$$ let t $ = tan\theta$

$2I= \int_0^{\infty}{ \frac{1}{(t+\sqrt{\frac{2}{4}})^2 + \frac{1}{2}} +\frac{1}{(t-\sqrt{\frac{2}{4}})^2 + \frac{1}{2}}$

$I= \frac{\sqrt{2}}{2}(tan^{-1}(\sqrt{2}t+1) +tan^{-1}(\sqrt{2}t-1))$

$= \frac{\pi \sqrt{2}}{2}$

porcupinetree · Jun 14, 2015

Re: MX2 2015 Integration Marathon

Drsoccerball said:
$I=\int_0^{\frac{\pi}{2}}{\frac{dx}{sin^4x +cos^4x}} = \frac{1}{2} \int_0^{\pi/2}{\frac{1}{sin^2x +cos^2x +\sqrt{2}sinxcosx}+ \frac{1}{sin^2x +cos^2x -\sqrt{2} cosxsinx}}$

$$ let t $ = tan\theta$

$2I= \int_0^{\infty}{ \frac{1}{(t+\sqrt{\frac{2}{4}})^2 + \frac{1}{2}} +\frac{1}{(t-\sqrt{\frac{2}{4}})^2 + \frac{1}{2}}$

$I= \frac{\sqrt{2}}{2}(tan^{-1}(\sqrt{2}t+1) +tan^{-1}(\sqrt{2}t-1))$

$= \frac{\pi \sqrt{2}}{2}$

Thanks. That's quite interesting how you change the limits from 0 to infinity - it never would have occurred to me that this was even allowed, but it makes sense

Drsoccerball · Jun 14, 2015

Re: MX2 2015 Integration Marathon

$\int{\frac{dx}{1-e^{kx}}}$

porcupinetree · Jun 14, 2015

Re: MX2 2015 Integration Marathon

Drsoccerball said:
$\int{\frac{dx}{1-e^{kx}}}$

https://imgur.com/lNE6F3Q

Carrotsticks · Jun 14, 2015

Re: MX2 2015 Integration Marathon

porcupinetree said:
https://imgur.com/lNE6F3Q

Another way which may be a little faster is to divide top and bottom by e^(kx) and it's a log integral.

Carrotsticks · Jun 14, 2015

Re: MX2 2015 Integration Marathon

porcupinetree said:
Thanks. That's quite interesting how you change the limits from 0 to infinity - it never would have occurred to me that this was even allowed, but it makes sense

You can turn the upper limit into T and then let T appraoch infinity, but it works pretty much the same way here.

Drsoccerball · Jun 14, 2015

Re: MX2 2015 Integration Marathon

porcupinetree said:
https://imgur.com/lNE6F3Q

OMG didnt see this D: i let x = lnu then multiplied top and bottom by u^(n-1) and got an integral which gives me your answerr:/
For this question let u=3+x and multiply top and bottom by $\sqrt{u-1}$ And the resulting integral is easily integratable

Carrotsticks · Jun 14, 2015

Re: MX2 2015 Integration Marathon

Drsoccerball said:
For : $\int_0^1{\frac{1-x^2}{(1+x^2)\sqrt{1+x^4}}}$

$$ Divide top and bottom by $ x^2 $ and let $ u=\frac{1}{x} +x$

Letting u=x^2+1/x^2 also works nicely.

Ekman · Jun 14, 2015

Re: MX2 2015 Integration Marathon

Carrotsticks said:
Another way which may be a little faster is to divide top and bottom by e^(kx) and it's a log integral.

Or you can sub in u=e^(kx) and use partial fractions.

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