HSC 2012-14 MX2 Integration Marathon (archive) (2 Viewers)

Sy123 · Sep 30, 2013

Re: MX2 Integration Marathon

$\int_{a}^b (x-a)^m (x-b)^n \ dx$

Sy123 · Sep 30, 2013

Re: MX2 Integration Marathon

$$i)$ \ \ I_n = \int_{-1}^0 \frac{x^n}{\sqrt{x+1}} \ dx$

$ii) Find in the form of an integral$

$\sum_{n=0}^{\infty} \frac{(-1)^n}{ \binom{2n}{n}}$

$iii) Evaluate this integral using Wolfram Alpha (or suffer evaluating it), note that$

$\tanh^{-1} x = \frac{1}{2} \ln \left |\frac{1+x}{1-x} \right |$

$Hence evaluate the sum$

Sy123 · Oct 10, 2013

Re: MX2 Integration Marathon

$\int_0^1 \sqrt{\frac{1}{x} - 1} \ dx$

JJ345 · Oct 10, 2013

Re: MX2 Integration Marathon

Sy123 said:
$\int_0^1 \sqrt{\frac{1}{x} - 1} \ dx$

Is it pi/2 ??
Surprisingly neat after such ugly working out.

Sy123 · Oct 10, 2013

Re: MX2 Integration Marathon

JJ345 said:
Is it pi/2 ??
Surprisingly neat after such ugly working out.

Yep

Drongoski · Oct 10, 2013

Re: MX2 Integration Marathon

Just saw this.

$\int ^1 _0 \sqrt {\frac {1}{x} - 1} dx = \int _0 ^1 \sqrt{\frac {1-x}{x}} dx = \int ^1 _0 \sqrt {\frac {1-x}{x}} \sqrt {\frac {1-x}{1-x}} dx = \int ^1 _0 \frac {1-x}{\sqrt {x-x^2}} dx \\ \\ = \frac {1}{2} \int ^1 _0 \frac {1-2x}{\sqrt {x-x^2}} + \frac {1}{\sqrt {(\frac {1}{2})^2 - (x- \frac {1}{2})^2}}dx \\ \\ = \frac {1}{2} \left [ 2\sqrt {x-x^2} + sin^{-1} (\frac {x-\frac{1}{2}}{\frac {1}{2}})\right]_0 ^1 \\ \\ = \frac {1}{2} \left [ 2 \sqrt{x-x^2} + sin^{-1}(2x-1)\right ] ^1 _0 \\ \\ = \frac {\pi}{2}$

Sy123 · Oct 10, 2013

Re: MX2 Integration Marathon

Drongoski said:
Just saw this.

$= \int _0 ^1 \sqrt{\frac {1-x}{x}} dx = \int ^1 _0 \sqrt {\frac {1-x}{x}} \sqrt {\frac {1-x}{1-x}} dx = \int ^1 _0 \frac {1-x}{\sqrt {x-x^2}} dx \\ \\ = \frac {1}{2} \int ^1 _0 \frac {1-2x}{\sqrt {x-x^2}} + \frac {1}{\sqrt {(\frac {1}{2})^2 - (x- \frac {1}{2})^2}}dx \\ \\ = \frac {1}{2} \left [ 2\sqrt {x-x^2} + sin^{-1} (\frac {x-\frac{1}{2}}{\frac {1}{2}})\right]_0 ^1 \\ \\ = \frac {\pi}{2} ??$

Interesting method, alternatively with the substitution, $x=\sin^2 \theta$ , it ends up being the integral of (sin^2theta)

Sy123 · Oct 13, 2013

Re: MX2 Integration Marathon

$\int \frac{\sin^2 x}{\cos^2 x + \cot^2 x} \ dx$

Carrotsticks · Oct 13, 2013

Re: MX2 Integration Marathon

Drongoski said:
Just saw this.

$\int ^1 _0 \sqrt {\frac {1}{x} - 1} dx = \int _0 ^1 \sqrt{\frac {1-x}{x}} dx = \int ^1 _0 \sqrt {\frac {1-x}{x}} \sqrt {\frac {1-x}{1-x}} dx = \int ^1 _0 \frac {1-x}{\sqrt {x-x^2}} dx \\ \\ = \frac {1}{2} \int ^1 _0 \frac {1-2x}{\sqrt {x-x^2}} + \frac {1}{\sqrt {(\frac {1}{2})^2 - (x- \frac {1}{2})^2}}dx \\ \\ = \frac {1}{2} \left [ 2\sqrt {x-x^2} + sin^{-1} (\frac {x-\frac{1}{2}}{\frac {1}{2}})\right]_0 ^1 \\ \\ = \frac {1}{2} \left [ 2 \sqrt{x-x^2} + sin^{-1}(2x-1)\right ] ^1 _0 \\ \\ = \frac {\pi}{2}$

Very neat result.

juantheron · Oct 14, 2013

Re: MX2 Integration Marathon

Sy123 said:
$\int \frac{\sin^2 x}{\cos^2 x + \cot^2 x} \ dx$

$\displaystyle \bf{=\int\frac{\sin^2 x}{\cos^2 x \cdot \left(1+\sec^2 x\cdot \frac{1}{\tan^2 x}\right)}dx}$

$\displaystyle \bf{ = \int\frac{\tan^{4}x}{\tan^2 x+1+\tan^2 x}dx = \int\frac{\tan^4 x}{2\tan^2 x+1}dx}$

$\displaystyle \bf{Now\;\; Let\;\; \tan x = t \Leftrightarrow \sec^2 xdx = dt\Rightarrow dx = \frac{1}{1+t^2}dt}$

$\displaystyle \bf{So\;\; Integration \;\; is \int\frac{t^4}{(2t^2+1)\cdot(t^2+1)}dt = \int \left(\frac{t^4.\left(2(t^2+1)-(2t^2+1)\right)}{(2t^2+1)\cdot(t^2+1)}\right)dt}$

$\displaystyle \bf{= \int \frac{2t^4}{2t^2+1}dt - \int\frac{t^4}{t^2+1}dt}$

$\displaystyle \bf{ = \frac{1}{2}.\int\frac{(2t^2)^2-1+1}{2t^2+1} - \int\frac{(t^2)^2-1+1}{t^2+1}dt}$

$\displaystyle \bf{ = \frac{1}{2}\int (2t^2-1)dt+\frac{1}{4}\int\frac{1}{t^2+(\sqrt{0.5})^2}dt - \int (t^2-1)dt-\int\frac{1}{t^2+1}dt}$

$\displaystyle \bf{ = \frac{1}{2}\left(\frac{2}{3}t^3-t\right)+\frac{1}{4\sqrt{0.5}}\tan^{-1}\left(\frac{t}{\sqrt{0.5}}\right)-\left(\frac{t^3}{3}-t\right)-\tan^{-1}(t)+\mathbb{C}}$

$\bf{where\;\; t = \tan x}$

juantheron · Oct 14, 2013

Re: MX2 Integration Marathon

$\displaystyle \bf{\int\frac{1}{(x^2-x+1)\sqrt{x^2+x+1}}dx}$

$\displaystyle \bf{ \int\sqrt{\frac{1+\tan x}{\csc^2 x+\sqrt{\sec x}}}dx}$

Sy123 · Oct 16, 2013

Re: MX2 Integration Marathon

juantheron said:
$\displaystyle \bf{=\int\frac{\sin^2 x}{\cos^2 x \cdot \left(1+\sec^2 x\cdot \frac{1}{\tan^2 x}\right)}dx}$

$\displaystyle \bf{ = \int\frac{\tan^{4}x}{\tan^2 x+1+\tan^2 x}dx = \int\frac{\tan^4 x}{2\tan^2 x+1}dx}$

$\displaystyle \bf{Now\;\; Let\;\; \tan x = t \Leftrightarrow \sec^2 xdx = dt\Rightarrow dx = \frac{1}{1+t^2}dt}$

$\displaystyle \bf{So\;\; Integration \;\; is \int\frac{t^4}{(2t^2+1)\cdot(t^2+1)}dt = \int \left(\frac{t^4.\left(2(t^2+1)-(2t^2+1)\right)}{(2t^2+1)\cdot(t^2+1)}\right)dt}$

$\displaystyle \bf{= \int \frac{2t^4}{2t^2+1}dt - \int\frac{t^4}{t^2+1}dt}$

$\displaystyle \bf{ = \frac{1}{2}.\int\frac{(2t^2)^2-1+1}{2t^2+1} - \int\frac{(t^2)^2-1+1}{t^2+1}dt}$

$\displaystyle \bf{ = \frac{1}{2}\int (2t^2-1)dt+\frac{1}{4}\int\frac{1}{t^2+(\sqrt{0.5})^2}dt - \int (t^2-1)dt-\int\frac{1}{t^2+1}dt}$

$\displaystyle \bf{ = \frac{1}{2}\left(\frac{2}{3}t^3-t\right)+\frac{1}{4\sqrt{0.5}}\tan^{-1}\left(\frac{t}{\sqrt{0.5}}\right)-\left(\frac{t^3}{3}-t\right)-\tan^{-1}(t)+\mathbb{C}}$

$\bf{where\;\; t = \tan x}$

Yea, well done
Alternatively:

$\int \frac{\sin^2 x}{\cos^2 x + \cot^2 x} \ dx = \int \frac{1+ \sin^4 x -1 }{\cos^2 x (1+\sin^2 x)} \ dx$

$= \int \frac{1+ \sin^4 x}{\cos^2 x (1+\sin^2 x)} \ dx - \int \frac{1}{\cos^2 x (1+\sin^2 x)} \ dx = \int \frac{\sin^2 x}{\cos^2 x} + \frac{1}{1+\sin^2 x} \ dx - \int \frac{1}{\cos^2x (1+\sin^2x)} \ dx$

$I_1 = \int \frac{\sin^2 x}{\cos^2 x} \ dx = \tan x - x$

$I_2 = \int \frac{1}{1+ \sin^2 x} \ dx = \int \frac{\sec^2x }{\sec^2 x + \tan^2 x} \ dx = \int \frac{\sec^2 x \ dx}{2\tan^2x + 1} = \frac{1}{\sqrt{2}} \tan^{-1} (\sqrt{2} \tan x)$

$I_3 = \int \frac{1}{\cos^2 x (1+\sin^2 x)} \ dx = \frac{1}{2} \int \frac{1}{\cos^2 x} + \frac{1}{1+\sin^2 x} \ dx = \frac{1}{2} \tan x + \frac{1}{2} I_2$

juantheron said:
$\displaystyle \bf{\int\frac{1}{(x^2-x+1)\sqrt{x^2+x+1}}dx}$

Can you please reveal the answer to this one?
There should be a more elegant way to do this than trig sub then half angle sub to yield a rational function

Sy123 · Oct 16, 2013

Re: MX2 Integration Marathon

$\int_0^1 \frac{x^{2012}}{1+x^2} \ dx$

HeroicPandas · Oct 16, 2013

Re: MX2 Integration Marathon

Sy123 said:
$\int_0^1 \frac{x^{2012}}{1+x^2} \ dx$

Generalise:

Let $I_n = \int_0^1 \frac{x^{2n}}{1+x^2}dx$

$I_n = \int_0^1\frac{x^{2n} + 1 -1}{1+x^2}dx \\ I_n = \left ( \int_0^1 \frac{(x^2)^n + 1}{1+x^2} dx \right )- \frac{\pi}{4} \\ \\ I_n = -\frac{\pi}{4} + \int_0^1 \frac{(1+x^2)(1 - x^2 + x^4 - \dots +(-1)^{n-1} (x^2)^{n-1})}{1+x^2}dx \\ \\ I_n = -\frac{\pi}{4} + \sum_{k = 1}^{n} (-1)^{k-1} \frac{x^{2k-1}}{2k-1} \\ \\ \therefore I_{2012} = -\frac{\pi}{4} + \sum_{k = 1}^{2012} (-1)^{k-1} \frac{x^{2k-1}}{2k-1}$ ?

ColdLipstick · Oct 16, 2013

Re: MX2 Integration Marathon

Sy123 said:
$\int_0^1 \frac{x^{2012}}{1+x^2} \ dx$

Not too certain about this answer... I'm new here.
Please correct me if it's wrong.

Sy123 · Oct 16, 2013

Re: MX2 Integration Marathon

HeroicPandas said:
Generalise:

Let $I_n = \int_0^1 \frac{x^{2n}}{1+x^2}dx$

$I_n = \int_0^1\frac{x^{2n} + 1 -1}{1+x^2}dx \\ I_n = \left ( \int_0^1 \frac{(x^2)^n + 1}{1+x^2} dx \right )- \frac{\pi}{4} \\ \\ I_n = -\frac{\pi}{4} + \int_0^1 \frac{(1+x^2)(1 - x^2 + x^4 - \dots +(-1)^{n-1} (x^2)^{n-1})}{1+x^2}dx \\ \\ I_n = -\frac{\pi}{4} + \sum_{k = 1}^{n} (-1)^{k-1} \frac{x^{2k-1}}{2k-1} \\ \\ \therefore I_{2012} = -\frac{\pi}{4} + \sum_{k = 1}^{2012} (-1)^{k-1} \frac{x^{2k-1}}{2k-1}$ ?

ColdLipstick said:
Not too certain about this answer... I'm new here.
Please correct me if it's wrong.

Yea something like that

well done

petermik · Oct 16, 2013

Re: MX2 Integration Marathon

twinklegal19 · Oct 16, 2013

Re: MX2 Integration Marathon

petermik said:
View attachment 28792

let x = tan@

you get end up with the integral of sec^5@

I'm not sure how to integrate it without using the recurrence formula?

Carrotsticks · Oct 16, 2013

Re: MX2 Integration Marathon

twinklegal19 said:
let x = tan@

you get end up with the integral of sec^5@

I'm not sure how to integrate it without using the recurrence formula?

Use sec^2 = 1+tan^2 and IBP.

twinklegal19 · Oct 16, 2013

Re: MX2 Integration Marathon

Carrotsticks said:
Use sec^2 = 1+tan^2 and IBP.

ah oops that's right lol

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