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

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SpiralFlex

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Re: MX2 Integration Marathon

1) Use the substitution x=sin^2u
2)(1+2(x(1-x))^.5)^.5 then becomes (1+2cosusinu)^.5
3)We the use the fact that sin^2u+cos^2u=1 to make the above square root into sinu+cosu
4) Then the integral becomes 2sinucos^2u +2sin^2ucosu
5) this becomes (2sin^3u)/3 - (2cos^3u)/3
6) this the becomes (2(x)^1.5)/3 - (2(1-x)^1.5)/3
One-two line method.
 

ForOneEon

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Re: MX2 Integration Marathon

I give up, my brain has been squished! Can you please your one-two line method? :D
Nooo! If your brain is squished, how will you do well in your exam?! Then again, my brain exploded trying to find the two-line method.
Explosion > Squished

So, Spiral. If you may please guide us to this two-line solution, we will be very grateful!
 
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rural juror

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Re: MX2 Integration Marathon

1) Use the substitution x=sin^2u
2)(1+2(x(1-x))^.5)^.5 then becomes (1+2cosusinu)^.5
3)We the use the fact that sin^2u+cos^2u=1 to make the above square root into sinu+cosu
4) Then the integral becomes 2sinucos^2u +2sin^2ucosu
5) this becomes (2sin^3u)/3 - (2cos^3u)/3
6) this the becomes (2(x)^1.5)/3 - (2(1-x)^1.5)/3

i subbed u = x-x^2 in,

solved the quadratic for x (take positvie root, its the same either way i think),
and cancelled difference of two sqaures and solved it throguh. got different answer -1/3(1-2u)^3/2 + (1-2u)^1/2

it might be the same expressiont hough
 

SpiralFlex

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Re: MX2 Integration Marathon

Nooo! If your brain is squished, how will you do well in your exam?! Then again, my brain exploded trying to find the two-line method.
Explosion > Squished

So, Spiral. If you may please guide us to this two-line solution, we will be very grateful!
I tricked you guys, this is a simple integral that requires the observation: (Level 1 question ;p)

 
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Sy123

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Re: MX2 Integration Marathon

 
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