HSC 2012-14 MX2 Integration Marathon (archive) (1 Viewer)

billym · Sep 5, 2013

Re: MX2 Integration Marathon

Sy123 said:
Yep well done

$\int_{-\infty}^{\infty} \frac{dx}{(x^2+a^2)(x^2+b^2)}$

Nice question. Is it pi/[ab(a+b)] ?

Sy123 · Sep 5, 2013

Re: MX2 Integration Marathon

billym said:
Nice question. Is it pi/[ab(a+b)] ?

Yep

integral95 · Sep 5, 2013

Re: MX2 Integration Marathon

billym said:
Nice question. Is it pi/[ab(a+b)] ?

Lol what steps did you take?

Sy123 · Sep 5, 2013

Re: MX2 Integration Marathon

$Evaluate$

$\lim_{n \to \infty} n \int_0^1 x^{n} \sqrt{x^2+1} \ dx$

(you may assume the limit converges)

asianese · Sep 6, 2013

MethewYan said:
Lol what steps did you take?

Partial fractions.

Sy123 · Sep 6, 2013

Re: MX2 Integration Marathon

$Using squeeze theorem$

$$Prove$ \ \ \lim_{n \to \infty} \frac{1}{n} \int_1^2 \left(1 - \frac{1}{x} \right)^n \ dx = 0$

Sy123 · Sep 8, 2013

Re: MX2 Integration Marathon

$\int_0^1 x^3 \sqrt{1-x^2} \ dx$

HeroicPandas · Sep 8, 2013

Re: MX2 Integration Marathon

Sy123 said:
$\int_0^1 x^3 \sqrt{1-x^2} \ dx$

$$let u^2 = 1-x^2, du = -xdx \\ \\ \therefore I = $\int_0^1 (1-u^2)u^2du = \left ( \frac{1}{3}- \frac{1}{5} \right ) = \frac{2}{15}$

Makematics · Sep 8, 2013

Re: MX2 Integration Marathon

HeroicPandas said:
$$let u^2 = 1-x^2, du = -xdx \\ \\ \therefore I = $\int_0^1 (1-u^2)u^2du = \left ( \frac{1}{3}- \frac{1}{5} \right ) = \frac{2}{15}$

OR you could do

$\\$Let $x=\sin\theta\\\\dx=\cos\theta~d\theta\\\\\sqrt{1-x^2}=\cos\theta\\\\x=0,~\theta=0\\\\x=1,~\theta= \frac{\pi}{2}\\\\\int_0^1 x^3 \sqrt{1-x^2} \ dx=\int_0^\frac{\pi}{2}\sin^3\theta\cos^2\theta~ d \theta \\\\~~~~~~~~~~~~~~~~~~~~~~~= -\int_0^\frac{\pi}{2}\cos^2\theta(1-\cos^2\theta)(-\sin\theta)~d\theta\\\\~~~~~~~~~~~~~~~~~~~~~~~=-\int_0^\frac{\pi}{2}(\cos^2\theta-\cos^4\theta)(-\sin\theta)~d\theta$
$\\$Let $u=\cos\theta\\\\du=-\sin\theta~d\theta\\\\\theta=0,~u=1\\\\\theta= \frac{\pi}{2},~u=0\\\\\int_0^1 x^3 \sqrt{1-x^2} \ dx=-\int_0^\frac{\pi}{2}(\cos^2\theta-\cos^4\theta)(-\sin\theta)~d\theta\\\\~~~~~~~~~~~~~~~~~~~~~~~=-\int_1^0(u^2-u^4)du\\\\~~~~~~~~~~~~~~~~~~~~~~~=\int_0^1(u^2-u^4)du\\\\~~~~~~~~~~~~~~~~~~~~~~~=[\frac{1}{3}u^3-\frac{1}{5}u^5]_0^1\\\\~~~~~~~~~~~~~~~~~~~~~~~=\frac{1}{3}-\frac{1}{5}\\\\~~~~~~~~~~~~~~~~~~~~~~~=\frac{2}{15}$

Your way is way better though

HeroicPandas · Sep 8, 2013

Re: MX2 Integration Marathon

Makematics said:
Your way is way better though

nahh its just the length, good effort man! in the end its all about answering the question

Sy123 · Sep 8, 2013

Re: MX2 Integration Marathon

Alternatively we do IBP with u=x^2, dv = xsqrt(1-x^2) dx

==

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

$y=\cot^{-1} x \ \ $is the inverse function of$ \ y=\cot x \ \ $for$ \ 0\leq x \leq \pi$

asianese · Sep 9, 2013

You should probably give a definition of inverse cotangent since it isn't explicitly defined in the syllabus.

Carrotsticks · Sep 9, 2013

Re: MX2 Integration Marathon

Sy123 said:
$\int_0^1 x^3 \sqrt{1-x^2} \ dx$

Try doing this one using no trig substitutions, and only the reverse chain rule.

I have a thing for solving complex problems using basic tools.

Sy123 · Sep 9, 2013

Re: MX2 Integration Marathon

Carrotsticks said:
Try doing this one using no trig substitutions, and only the reverse chain rule.

I have a thing for solving complex problems using basic tools.

Good idea

$x^3\sqrt{1-x^2} = -x\sqrt{1-x^2} (1-x^2-1) = x\sqrt{1-x^2} - x(1-x^2)^{\frac{3}{2}}$

Which can be done using reverse chain rule

What makes integration one of my favorite topics is the vast amount of ways to get to the same answer, I think we've all shown up to 4 ways to the answer so far

Sy123 · Sep 9, 2013

Re: MX2 Integration Marathon

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

levaly · Sep 11, 2013

Re: MX2 Integration Marathon

Makematics said:
OR you could do

$\\$Let $x=\sin\theta\\\\dx=\cos\theta~d\theta\\\\\sqrt{1-x^2}=\cos\theta\\\\x=0,~\theta=0\\\\x=1,~\theta= \frac{\pi}{2}\\\\\int_0^1 x^3 \sqrt{1-x^2} \ dx=\int_0^\frac{\pi}{2}\sin^3\theta\cos^2\theta~ d \theta \\\\~~~~~~~~~~~~~~~~~~~~~~~= -\int_0^\frac{\pi}{2}\cos^2\theta(1-\cos^2\theta)(-\sin\theta)~d\theta\\\\~~~~~~~~~~~~~~~~~~~~~~~=-\int_0^\frac{\pi}{2}(\cos^2\theta-\cos^4\theta)(-\sin\theta)~d\theta$
$\\$Let $u=\cos\theta\\\\du=-\sin\theta~d\theta\\\\\theta=0,~u=1\\\\\theta= \frac{\pi}{2},~u=0\\\\\int_0^1 x^3 \sqrt{1-x^2} \ dx=-\int_0^\frac{\pi}{2}(\cos^2\theta-\cos^4\theta)(-\sin\theta)~d\theta\\\\~~~~~~~~~~~~~~~~~~~~~~~=-\int_1^0(u^2-u^4)du\\\\~~~~~~~~~~~~~~~~~~~~~~~=\int_0^1(u^2-u^4)du\\\\~~~~~~~~~~~~~~~~~~~~~~~=[\frac{1}{3}u^3-\frac{1}{5}u^5]_0^1\\\\~~~~~~~~~~~~~~~~~~~~~~~=\frac{1}{3}-\frac{1}{5}\\\\~~~~~~~~~~~~~~~~~~~~~~~=\frac{2}{15}$

Your way is way better though

Here is another option for a solution:
http://www.symbolab.com/solutions/inequalities?query=\int_{0}^{1} x^{3}\sqrt{1-x^{2}}dx

braintic · Sep 11, 2013

Re: MX2 Integration Marathon

Please STOP using this site to promote your own stupid website.

integral95 · Sep 12, 2013

Re: MX2 Integration Marathon

Sy123 said:
$Using squeeze theorem$

$$Prove$ \ \ \lim_{n \to \infty} \frac{1}{n} \int_1^2 \left(1 - \frac{1}{x} \right)^n \ dx = 0$

for 1<x<2

1/2<(1-1/x)<1

0<(1-1/x)^n<1

Divide the whole inequality by n

Integrate the whole thing with respect to x with limits 0 to 1

0<(integral)<1/n
Then apply the limit as n approaches infinity to get the result

Makematics · Sep 13, 2013

Re: MX2 Integration Marathon

Sy123 said:
$\int \frac{x^2-1}{x(x^2+1)(\ln (x^2+1) - \ln x)} \ dx$

Great question

I noticed you love the ones that use u=x+1/x as a substitution haha

$\\\int \frac{x^2-1}{x(x^2+1)(\ln (x^2+1) - \ln x)} \ dx=\int\frac{x^2-1}{x(x^2+1)\ln{\frac{x^2+1}{x}}}~dx\\\\~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~=\int\frac{1-\frac{1}{x^2}}{(x+\frac{1}{x})\ln(x+\frac{1}{x})}~dx\\\\$Let $u=x+\frac{1}{x}\\\\du=(1-\frac{1}{x^2})~dx\\\\\int \frac{x^2-1}{x(x^2+1)(\ln (x^2+1) - \ln x)} \ dx=\int\frac{1-\frac{1}{x^2}}{(x+\frac{1}{x})\ln(x+\frac{1}{x})}~dx\\\\~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~=\int\frac{du}{u\ln{u}}\\\\~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~=\ln{(\ln{u})}+c\\\\~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~=\ln{(\ln{(x+\frac{1}{x})})}+c$

Sy123 · Sep 13, 2013

Re: MX2 Integration Marathon

Makematics said:
Great question I noticed you love the ones that use u=x+1/x as a substitution haha

$\\\int \frac{x^2-1}{x(x^2+1)(\ln (x^2+1) - \ln x)} \ dx=\int\frac{x^2-1}{x(x^2+1)\ln{\frac{x^2+1}{x}}}~dx\\\\~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~=\int\frac{1-\frac{1}{x^2}}{(x+\frac{1}{x})\ln(x+\frac{1}{x})}~dx\\\\$Let $u=x+\frac{1}{x}\\\\du=(1-\frac{1}{x^2})~dx\\\\\int \frac{x^2-1}{x(x^2+1)(\ln (x^2+1) - \ln x)} \ dx=\int\frac{1-\frac{1}{x^2}}{(x+\frac{1}{x})\ln(x+\frac{1}{x})}~dx\\\\~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~=\int\frac{du}{u\ln{u}}\\\\~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~=\ln{(\ln{u})}+c\\\\~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~=\ln{(\ln{(x+\frac{1}{x})})}+c$

Yeah I do quite like those substitutions, it seems so elegant.
Like those random ones people think of like x=(1-u)/(1+u) or whatever

$\int_0^{\pi /2} \frac{\sin(2013x)}{\sin x} \ dx$

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