HSC 2012-14 MX2 Integration Marathon (archive) (3 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.

HSC 2012-14 MX2 Integration Marathon (archive) (3 Viewers)

Well-Known Member

This too shall pass

Retired

This too shall pass

Premium Member

This too shall pass

Premium Member

Premium Member

what is that?It is Cowpea

Premium Member

what is that?It is Cowpea

Premium Member

Premium Member

what is that?It is Cowpea

This too shall pass

Premium Member

This too shall pass

Well-Known Member

Well-Known Member

Premium Member

Users Who Are Viewing This Thread (Users: 0, Guests: 3)