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

braintic · May 23, 2013

Re: MX2 Integration Marathon

Sy123 said:
The substitution (u^2-1)^2 can be modified slightly to make the integral much more easier. Its close though.

Actually the integral became just 4∫(u^4 - u^2) du - about as simple as it gets. The messy part was the ensuing substitution. I've checked my answer by differentiation and it works.

Sy123 · May 23, 2013

Re: MX2 Integration Marathon

braintic said:
Actually the integral became just 4∫(u^4 - u^2) du - about as simple as it gets. The messy part was the ensuing substitution. I've checked my answer by differentiation and it works.

Yep, I was thinking of (u-1)^2 which is slightly longer.

===========

$\int \tan^{-1} \left(\frac{1}{1+x(x-1)} \right ) \ dx$

Carrotsticks · May 23, 2013

Re: MX2 Integration Marathon

Personally, I would have made the substitution x=u^2.

As a result, you get the integral of u*sqrt(1+u), which is straight-forward to evaluate, and the substitution back to x is quick.

Sy123 · May 23, 2013

Re: MX2 Integration Marathon

$\int_0^1 x^5 \sqrt{1+x^3} \ dx$

anomalousdecay · May 23, 2013

Re: MX2 Integration Marathon

Sy123 said:
$\int_0^1 x^5 \sqrt{1+x^3} \ dx$

$\text{Let:} u= x^3, dv= \sqrt{(1 + x^3)}dx$

$du=3x^2 dx,\:v = \frac{2\sqrt{(1 +x^3)^3}}{9}$

$\int_0^1 x^5 \sqrt{1+x^3} \ dx$

$= (x^3) \left (\frac{2\sqrt{(1 +x^3)^3}}{9} \right) - \int_0^1 \left (\frac{2x^2\sqrt{(1+x^3)^3}}{3} \right ) dx$

$= \frac{2\sqrt{2}}{9} - \frac{5\sqrt{(1+x^3)^3}}{9}$

$= \frac{2\sqrt{2}}{9} - \frac{10\sqrt{2}}{9} + \frac{5}{9}$

$= \frac{5-8\sqrt{2}}{9}$

Just learnt Integration by parts and Latex so probably wrong.

Sy123 · May 23, 2013

Re: MX2 Integration Marathon

anomalousdecay said:
$\text{Let:} u= x^3, dv= \sqrt{(1 + x^3)}dx$

$du=3x^2 dx,\:v = \frac{2\sqrt{(1 +x^3)^3}}{9}$

$\int_0^1 x^5 \sqrt{1+x^3} \ dx$

$= (x^3) \left (\frac{2\sqrt{(1 +x^3)^3}}{9} \right) - \int_0^1 \left (\frac{2x^2\sqrt{(1+x^3)^3}}{3} \right ) dx$

$= \frac{2\sqrt{2}}{9} - \frac{5\sqrt{(1+x^3)^3}}{9}$

$= \frac{2\sqrt{2}}{9} - \frac{10\sqrt{2}}{9} + \frac{5}{9}$

$= \frac{5-8\sqrt{2}}{9}$

Just learnt Integration by parts and Latex so probably wrong.

Well yeah if its right its right, nice work.

Alternatively the substitution u=1+x^3 and the split of the first x^5 into x^3 and x^2 yields a faster method.

anomalousdecay · May 23, 2013

Re: MX2 Integration Marathon

Sy123 said:
Well yeah if its right its right, nice work.

Alternatively the substitution u=1+x^3 and the split of the first x^5 into x^3 and x^2 yields a faster method.

Yeah I did split it. u= x^2. It is pretty obvious with integration by parts though as you need to get rid of the coefficient as efficiently as possible.

I usually have problems more with partial fractions and splitting the numerator though.

I can't think of a really hard question atm.

anomalousdecay · May 23, 2013

Re: MX2 Integration Marathon

$\int_0^1 \left ({{cos^{-1}x} + {tan^{-1}x}} \right) dx$

I could not think of something creative at this point of time.

RealiseNothing · May 23, 2013

Re: MX2 Integration Marathon

anomalousdecay said:
$\int_0^1 \left ({{cos^{-1}x} + {tan^{-1}x}} \right) dx$

I could not think of something creative at this point of time.

What you can do is consider areas. First split the integral into:

$\int_{0}^{1} cos^{-1}x~dx + \int_{0}^{1} tan^{-1}x~dx$

Now if we consider the integrals separately and consider the areas bound by the curves:

$\int_{0}^{1} cos^{-1}x~dx = \int_{0}^{\frac{\pi}{2}} cosx~dx$

$=1$

Similarly for inverse tan:

$\int_{0}^{1} tan^{-1}x~dx = \frac{\pi}{4} - \int_{0}^{\frac{\pi}{4}} tanx~dx$

$=\frac{\pi}{4} - \frac{ln(2)}{2}$

Adding the two integrals together gives:

$1 + \frac{\pi}{4} - \frac{ln(2)}{2}$

anomalousdecay · May 23, 2013

Re: MX2 Integration Marathon

RealiseNothing said:
What you can do is consider areas. First split the integral into:

$\int_{0}^{1} cos^{-1}x~dx + \int_{0}^{1} tan^{-1}x~dx$

Now if we consider the integrals separately and consider the areas bound by the curves:

$\int_{0}^{1} cos^{-1}x~dx = \int_{0}^{\frac{\pi}{2}} cosx~dx$

$=1$

Similarly for inverse tan:

$\int_{0}^{1} tan^{-1}x~dx = \frac{\pi}{4} - \int_{0}^{\frac{\pi}{4}} tanx~dx$

$=\frac{\pi}{4} - \frac{ln(2)}{2}$

Adding the two integrals together gives:

$1 + \frac{\pi}{4} - \frac{ln(2)}{2}$

Edit: Never Mind you were right. Nice Work.

RealiseNothing · May 23, 2013

Re: MX2 Integration Marathon

anomalousdecay said:
I got this using the longer method:

$\frac{pi-4}{4} - ln\frac{\sqrt{2}{2}}$

$ln{\frac{\sqrt{2}}{2}} = ln{\frac{1}{\sqrt{2}}} = ln{2^{\frac{-1}{2}} = \frac{-ln{2}}{2}$

anomalousdecay · May 23, 2013

Re: MX2 Integration Marathon

RealiseNothing said:
What you can do is consider areas. First split the integral into:

$\int_{0}^{1} cos^{-1}x~dx + \int_{0}^{1} tan^{-1}x~dx$

Now if we consider the integrals separately and consider the areas bound by the curves:

$\int_{0}^{1} cos^{-1}x~dx = \int_{0}^{\frac{\pi}{2}} cosx~dx$

$=1$

Similarly for inverse tan:

$\int_{0}^{1} tan^{-1}x~dx = \frac{\pi}{4} - \int_{0}^{\frac{\pi}{4}} tanx~dx$

$=\frac{\pi}{4} - \frac{ln(2)}{2}$

Adding the two integrals together gives:

$1 + \frac{\pi}{4} - \frac{ln(2)}{2}$

Nice work. That is right and I got the same answer doing it the long way.

anomalousdecay · May 23, 2013

Re: MX2 Integration Marathon

RealiseNothing said:
$ln{\frac{\sqrt{2}}{2}} = ln{\frac{1}{\sqrt{2}}} = ln{2^{\frac{-1}{2}} = \frac{-ln{2}}{2}$

Yeah my bad. You can see the my LaTex fails in that lol

RealiseNothing · May 23, 2013

Re: MX2 Integration Marathon

anomalousdecay said:
Yeah my bad. You can see the my LaTex fails in that lol

Obviously haha.

I'll try come up with something interesting, I posted one in the 3U thread but it's not that difficult.

Sy123 · May 24, 2013

Re: MX2 Integration Marathon

$\int_{1/3}^{\infty} \frac{5\sqrt{x^2 - \frac{1}{9}} + 69x^6(x^2 - \frac{1}{9})}{x^6(x^2- \frac{1}{9})} \ dx$

I assure you the numbers are not arbitrary.

anomalousdecay · May 24, 2013

Re: MX2 Integration Marathon

Sy123 said:
$\int_{1/3}^{\infty} \frac{5\sqrt{x^2 - \frac{1}{9}} + 69x^6(x^2 - \frac{1}{9})}{x^6(x^2- \frac{1}{9})} \ dx$

I assure you the numbers are not arbitrary.

$\int^{\infty}_{\frac{1}{3}}{ \left( \frac{5}{x^6\sqrt{x^2-\frac{1}{9}}} + 69 \right)} dx$

$\frac{5}{x^6\sqrt{x^2-\frac{1}{9}}} = \frac{225}{x^{12}(3x-1)(3x+1)} = .....$

Used partial fractions but got freakishly large numbers (like 500000).
I am guessing you don't use partial fractions?

Sy123 · May 24, 2013

Re: MX2 Integration Marathon

anomalousdecay said:
$\int^{\infity}_{\frac{1}{3}}{ \left( \frac{5}{x^6\sqrt{x^2-\frac{1}{9}} + 69 \right)} dx$

$\frac{5}{x^6\sqrt{x^2-\frac{1}{9}}} = \frac{225}{x^(12)(3x-1)(3x+1)} = .....$

Used partial fractions but got freakishly large numbers (like 500000).
I am guessing you don't use partial fractions?

Yeah I didn't use partial fractions, I recommend a substitution.

Makematics · May 24, 2013

Re: MX2 Integration Marathon

doing partial fractions on that sounds worse than death LOL

Makematics · May 24, 2013

Re: MX2 Integration Marathon

hmmm for Sy's question i experimented with x=1/3sectheta but to no avail

was getting very close until something stuffed up with the limits :/ i got sec (pi/2)

anomalousdecay · May 24, 2013

Re: MX2 Integration Marathon

I think maybe tanx or cotx are best as they represent a way of eliminating infinity.

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