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

dunjaaa · Aug 21, 2014

Re: MX2 Integration Marathon

Screen shot 2014-08-21 at 4.18.39 PM.png

dunjaaa · Sep 9, 2014

Re: MX2 Integration Marathon

Screen shot 2014-09-09 at 7.53.13 PM.png

for n>0

dunjaaa · Sep 9, 2014

Re: MX2 Integration Marathon

Screen shot 2014-09-09 at 8.04.27 PM.png

Ikki · Sep 9, 2014

Re: MX2 Integration Marathon

dunjaaa said:
View attachment 30890

By inspection if we expand the bottom the power of x will be 5/6. Hence let u=x^1/6
Then the integral is reduced to 6u^2/(1+u^2), factor out constants, +1-1 and you get a 6 as a constant and a tan inverse integration (of 1=pi/4)...
=6*(1-pi/4)
=6-2pi/3

Nice shifty U-sub haha

Ikki · Sep 9, 2014

Re: MX2 Integration Marathon

dunjaaa said:
View attachment 30889 for n>0

This time let u=x^n+1, do some partial fractions, inside the log you can split into 1-1/u which when put under boundaries of 2 to infinity...
1/infinity=0
Hence = 1/n * [ln(1)-ln(1/2)]
=1/n * ln(1/2)

faisalabdul16 · Sep 9, 2014

Re: MX2 Integration Marathon

Ikki said:
By inspection if we expand the bottom the power of x will be 5/6. Hence let u=x^1/6
Then the integral is reduced to 6u^2/(1+u^2), factor out constants, +1-1 and you get a 6 as a constant and a tan inverse integration (of 1=pi/4)...
=6*(1-pi/4)
=6-2pi/3

Nice shifty U-sub haha

Nice !
I did two substitutions lol. Nevertheless got the same answer.

dunjaaa · Sep 9, 2014

Re: MX2 Integration Marathon

Nice work! Here's what I did: Factor x^n from the (1+x^n), then let u=1+1/x^n and it falls out nicely. Nice generalisation aye

Ikki · Sep 9, 2014

Re: MX2 Integration Marathon

Hehe made an easy q, What is this to 2 dp:
$e^(\int_{-1.518213265}^{-pi/2+0.0005005}\frac{2}{2cos^2x-sin2x}dx)$

Ikki · Sep 9, 2014

Re: MX2 Integration Marathon

Here's an actual challenging Q: (Well not for you guys ofcourse

)
$\int_{-1}^{1}\frac{2x}{(x^2+2x+5)^2}dx$

dunjaaa · Sep 9, 2014

Re: MX2 Integration Marathon

I remember doing this question in a trial paper

. I'll post a solution tomorrow if nobody has solved it yet. Hint: Use a trig-substitution

dunjaaa · Sep 9, 2014

Re: MX2 Integration Marathon

Screen shot 2014-09-09 at 9.59.51 PM.png

, what's with the decimals lol

integral95 · Sep 9, 2014

Re: MX2 Integration Marathon

Ikki said:
Hehe made an easy q, What is this to 2 dp:
$e^(\int_{-1.518213265}^{-pi/2+0.0005005}\frac{2}{2cos^2x-sin2x}dx)$

Well the hard part is the integral really so i'll just do that

$\frac{2}{2cos^2x-sin2x} = \frac{1}{cos^2x - sinxcosx} \\ \\ \\ = \frac{1}{cosx(cosx-sinx)} = \frac{sin^2x+cos^2x}{{cosx(cosx-sinx)}} = \frac{cosx +sinx}{cosx - sin x} - \frac{sinx}{cosx} \\ \\ \therefore \int\frac{2}{2cos^2x-sin2x} = \int \frac{cosx +sinx}{cosx - sin x} - \frac{sinx}{cosx} = -ln(cosx - sinx) + ln(cosx) \\ \\ \\ = ln(\frac{cosx}{cosx - sinx})$

integral95 · Sep 9, 2014

Re: MX2 Integration Marathon

Ikki said:
Here's an actual challenging Q: (Well not for you guys ofcourse )
$\int_{-1}^{1}\frac{2x}{(x^2+2x+5)^2}dx$

I wanna do it

$\int_{-1}^{1}\frac{2x}{(((x+1)^2)+4)^2} \\ \\ \\ $Let x+1 = tan\theta$ \\ \\ 2\int_{0}^\frac{\pi}{4}\frac{(2tan\theta-1)sec^2\theta}{16sec^4\theta} \\ \\ = \int_{0}^\frac{\pi}{4}\frac {2tan\theta-1}{8sec^2\theta} = \frac{1}{4}\int_{0}^\frac{\pi}{4}sin\theta cos\theta d\theta \ - \frac{1}{8}\int_{0}^{\frac{\pi}{4}} cos^2\theta d\theta \\ \\ = \frac{1}{8}\int_{0}^{\frac{\pi}{4}}sin2\theta -cos^2\theta \ d\theta \\ \\ = \frac{-1}{16}[cos2\theta + \theta + \frac{1}{2}sin2\theta] = \frac{1}{16} - \frac{\pi}{64}$

nightweaver066 · Sep 9, 2014

Re: MX2 Integration Marathon

Ikki said:
Hehe made an easy q, What is this to 2 dp:
$e^(\int_{-1.518213265}^{-pi/2+0.0005005}\frac{2}{2cos^2x-sin2x}dx)$

An easier way..
$\int \frac{1}{cos^2x-sinxcosx}dx=\int \frac{sec^2x}{1-tanx}dx = -ln|1-tanx|$

dunjaaa · Sep 9, 2014

Re: MX2 Integration Marathon

nice

Ikki · Sep 10, 2014

Re: MX2 Integration Marathon

Well done guys

Whoops @dunjaa switch the boundaries and remember that it's e to the power of the answer

dan964 · Sep 10, 2014

Re: MX2 Integration Marathon

Integrate e^(xe^(x^2))

Ikki · Sep 10, 2014

Re: MX2 Integration Marathon

dan964 said:
Integrate e^(xe^(x^2))

So that's e to the power of e to the power of e^x squared, hmm does that work?

Carrotsticks · Sep 10, 2014

Re: MX2 Integration Marathon

Ikki said:
So that's e to the power of e to the power of e^x squared, hmm does that work?

Yep, look up "Power Towers" in google.

dan964 · Sep 10, 2014

Re: MX2 Integration Marathon

Ikki said:
So that's e to the power of e to the power of e^x squared, hmm does that work?

No it is e to the power of x times e to the power of x^2

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

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