HSC 2012-14 MX2 Integration Marathon (archive) (4 Viewers)

twinklegal19 · May 1, 2013

asianese said:
Let's all get exotic with our constants! (what)

haha

Sy123 · May 1, 2013

Re: MX2 Integration Marathon

$Evaluate$

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

$$You may NOT use the substitution$ \ \ x=\frac{1-u}{1+u}$

RealiseNothing · May 1, 2013

Re: MX2 Integration Marathon

Sy123 said:
$Evaluate$

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

$$You may NOT use the substitution$ \ \ x=\frac{1-u}{1+u}$

Let $x = \frac{1-p}{1+p}$

trololololol

Sy123 · May 1, 2013

Re: MX2 Integration Marathon

RealiseNothing said:
Let $x = \frac{1-p}{1+p}$

trololololol

hehehe :s
I'd prefer someone find a solution without that substitution because that has already been posted in this thread.

Sy123 · May 2, 2013

Re: MX2 Integration Marathon

Sy123 said:
$I=\int \sqrt{\frac{e^x-1}{e^x+1}} \ dx$

$I = \int \frac{\sqrt{e^{2x}-1}}{e^x+1} \ dx$

$e^x = \sec u$

$e^x \ dx = \sec u \tan u \ du$

$dx=\tan u \ du$

$I = \int \frac{\tan^2 u}{\sec u +1} \ du$

$I = \int \sec u -1 \ du$

$I = \ln(\sec u + \tan u) - u$

$\therefore \ I = \cos^{-1} (e^{-x}) + \ln(e^x+\sqrt{e^{2x}-1}) + (asianese)$

Drongoski · May 2, 2013

Re: MX2 Integration Marathon

Very nice - Omar Khayyam!

yr typo: dx = tanu du

Sy123 · May 2, 2013

Re: MX2 Integration Marathon

Drongoski said:
Very nice - Omar Khayyam!

yr typo: dx = tanu du

Thank you
And its actually Ibn Sina

Capt Rifle · May 2, 2013

Re: MX2 Integration Marathon

seanieg89 said:
Not really, its just one branch.

Yes Dr.

asianese · May 2, 2013

Re: MX2 Integration Marathon

Dr to be

Examine · May 2, 2013

Re: MX2 Integration Marathon

$\\ $Consider the curve $ y=\frac{1}{1+x^2} $ for all $ x\geq 1 \\\\ $An arbitrary point on the curve $ P(x,y) $ is chosen and a line is drawn from $ P $ to the $ y $ intercept, forming a trapezium with the $ x $ axis.$ \\\\ $This trapezium under-estimates the area under the curve for some values of $ x $ and over-estimates the area under the curve for other values of $ x. $ Find the value of $ x $ (or form an equation whose solution is that value of $ x $) such that the area of the trapezium is equal to the area under the curve.$$

Sy123 · May 2, 2013

Re: MX2 Integration Marathon

Examine said:
$\\ $Consider the curve $ y=\frac{1}{1+x^2} $ for all $ x\geq 1 \\\\ $An arbitrary point on the curve $ P(x,y) $ is chosen and a line is drawn from $ P $ to the $ y $ intercept, forming a trapezium with the $ x $ axis.$ \\\\ $This trapezium under-estimates the area under the curve for some values of $ x $ and over-estimates the area under the curve for other values of $ x. $ Find the value of $ x $ (or form an equation whose solution is that value of $ x $) such that the area of the trapezium is equal to the area under the curve.$$

The question is missing some information, but I'm assuming the facts you left out were:
1) Draw a line from P perpendicular to the x-axis, and that forms the trapezium?

2) When you mean area underneath the curve, I think you mean:

$A = \int_0^{x} \frac{1}{1+t^2} \ dt$ And how well the trapezium estimates this (where P(t, y(t)) )

Is that what you meant?

I'll leave this for someone else to do though

Sy123 · May 4, 2013

Re: MX2 Integration Marathon

Sy123 said:
$Evaluate$

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

$$You may NOT use the substitution$ \ \ x=\frac{1-u}{1+u}$

$x=\tan \theta$

$I = \int_0^{\pi/4} ln(1+ \tan \theta) \ d\theta$

$\int_0^{a} f(x) \ dx = \int_0^a f(a-x) \ dx$

$\int_0^{\pi/4} \ln(1+\tan \theta) \ d\theta = \int_0^{\pi/4} \ln \frac{2}{1+ \tan \theta} \ d\theta$

$2I = \int_0^{\pi/4} \ln(2)$

$I = \frac{\pi}{8} \ln 2$

=================

$Find$

$\int \frac{\sin^{-1}(e^x)}{e^x} \ dx$

Registered User · May 4, 2013

Re: MX2 Integration Marathon

Sy123 said:
$Evaluate$

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

$$You may NOT use the substitution$ \ \ x=\frac{1-u}{1+u}$

I think this same question is on the 4U marathon thread

The tan sub is easier and more obvious, isn't it?

Sy123 · May 4, 2013

Re: MX2 Integration Marathon

Registered User said:
I think this same question is on the 4U marathon thread
The tan sub is easier and more obvious, isn't it?

Is it?

Well I asked for the question to be done without that particular substitution because earlier in this Integration marathon someone already solved it using that substitution.
Tan substitution is a bit obvious, though the definite integrals property I think is less obvious.

Registered User · May 4, 2013

Re: MX2 Integration Marathon

Sy123 said:
$Find$

$\int \frac{\sin^{-1}(e^x)}{e^x} \ dx$

Not sure if this is correct because it looks really ugly...

$Let$ $I=\int \frac{sin^{-1}(e^{x})}{e^{x}}$

$Let \frac{1}{e^{x}}=u\therefore dx=-due^{x}$

$=-\int sin^{-1}(\frac{1}{u})du$

$Let \frac{1}{u}=v\therefore du=-\frac{dv}{v^{2}}$

$\therefore I=\int \frac{sin^{-1}(v)}{v^{2}}dv$

$Now do integration by parts. Let$ $u=sin^{-1}(v)$ $and$ $dv=\frac{1}{v^{2}} \cdots$

$\Therefore I=-\frac{\sqrt{1-(e^{x})^{2}}\frac{1}{e^{x}}ln( \sqrt {\frac{1}{e^{x}}^{2}-1})+\frac{1}{e^{x}}}{\sqrt{\frac{1}{e^{x}}^{2}-1})}-\frac{1}{e^{x}}sin^{-1}(e^{x})+C$

Registered User · May 4, 2013

Re: MX2 Integration Marathon

Sy123 said:
Is it?

Well I asked for the question to be done without that particular substitution because earlier in this Integration marathon someone already solved it using that substitution.
Tan substitution is a bit obvious, though the definite integrals property I think is less obvious.

Yeah page 14.
No one actually did it though but people mentioned the two methods.

Sy123 · May 4, 2013

Re: MX2 Integration Marathon

Registered User said:
Not sure if this is correct because it looks really ugly...

$Let$ $I=\int \frac{sin^{-1}(e^{x})}{e^{x}}$

$Let \frac{1}{e^{x}}=u\therefore dx=-due^{x}$

$=-\int sin^{-1}(\frac{1}{u})du$

$Let \frac{1}{u}=v\therefore du=-\frac{dv}{v^{2}}$

$\therefore I=\int \frac{sin^{-1}(v)}{v^{2}}dv$

$Now do integration by parts. Let$ $u=sin^{-1}(v)$ $and$ $dv=\frac{1}{v^{2}} \cdots$

$\Therefore I=-\frac{\sqrt{1-(e^{x})^{2}}\frac{1}{e^{x}}ln( \sqrt {\frac{1}{e^{x}}^{2}-1})+\frac{1}{e^{x}}}{\sqrt{\frac{1}{e^{x}}^{2}-1})}-\frac{1}{e^{x}}sin^{-1}(e^{x})+C$

You may have the answer but in a different form.
But you have the idea, its just by parts then a substitution on the second integral.

===================

$\int \frac{1}{x^{n} + x} \ dx$

Registered User · May 4, 2013

Re: MX2 Integration Marathon

Sy123 said:
$\int \frac{1}{x^{n} + x} \ dx$

Nice question.

Looks hard but can be easily done using a substitution.

seanieg89 · May 4, 2013

Re: MX2 Integration Marathon

Registered User said:
Nice question.

Looks hard but can be easily done using a substitution.

We don't even need substitution:

$I=\int \frac{1}{x^n+x} \, dx=\int \frac{1}{x}-\frac{x^{n-2}}{x^{n-1}+1} \, dx=\log |x|-\frac{1}{n-1}\log |x^{n-1}+1|+C$

Sy123 · May 4, 2013

Re: MX2 Integration Marathon

$I_n = \int_{0}^{\frac{\pi}{2}} \cos^{2n} x \ dx$

$Prove that$

$\sum_{n=1}^{\infty} \frac{I_n}{2n-1} = \frac{\pi}{2}$

HSC 2012-14 MX2 Integration Marathon (archive) (4 Viewers)

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