integration question (1 Viewer)

[SaHbEeWaH]

hi can somebody show me how to do this

integrate x^6 / sqrt(1 - x^2)

thanks

 

[Jago]

i've been told this integration is more 4unit orientated. try asking the question there.

 

[Poon-Tang]

x to the 6 integrates to be x to the 7 over 7
and all that is over
the integral of root(1-x cubed) which is (1-xcubed) to the half
which integral is 1-xcubed to the 1.5 over 1.5 times the differentail of whats inside the brackets (2x)

so:
x (to the 7)
over 7
over (1-xcubed) to the 1.5
over 1.5 times 2x

 

[Poon-Tang]

ps. i do 2 unit maths so i'm probably wrong

 

[Estel]

You can't quite integrate that way Poon
It's a simple substitution with x = cosa, or x = sina (up to your whim).

 

[Slidey]

x to the 6 integrates to be x to the 7 over 7
and all that is over
the integral of root(1-x cubed) which is (1-xcubed) to the half
which integral is 1-xcubed to the 1.5 over 1.5 times the differentail of whats inside the brackets (2x)

so:
x (to the 7)
over 7
over (1-xcubed) to the 1.5
over 1.5 times 2x

Unfortunately, that is essentially treating the integral of g(x)*h(x) as the integral of g(x) * the integral of h(x). This is not possible.

I=Int( x^6/sqrt(1 - x^2) )dx
Let x=sin@ (I like this one because the derivative is positive)
dx=cos@ d@
I=Int( sin^6(@)/cos@ * cos@ d@)
I=Int( sin^6(@) d@ )

I'm about to use a 4 unit technique, so you should probably stop reading.

Now, 2isin@=z-1/z
(2isin@)^6=-64sin^6(@)=(z-1/z)^6
1 6 15 20 15 6 1
-64sin^6(@)=(z^6+1/z^6)+6(z^4+1/z^4)+15(z^2+1/z^2)+20
.'. subbing back in z^n+1/z^n=2cos(n@):
-64sin^6(@)=2cos6@+12cos4@+30cos2@+20
.'.sin^6(@)=-(cos6@+6cos4@+15cos2@+10)/32

So I=Int(sin^6(@) d@)
Equals I=Int(-(cos6@+6cos4@+15cos2@+10)/32)
I=-Int(cos6@+6cos4@+15cos2@+10)/32
I=-[sin(6@)/6+2sin(4@)/3+15sin(2@)/2+10@]/32

I=-(sin6@+4sin4@+45sin2@+60@)/192

Or something like that. Kinda bored.

 

[Jago]

my eyes sorta went out of focus due to all the "@"s...

S.R. : show me how to do induction inequalities properly in the other thread.

 

[Estel]

rofl at your method Slide Rule!
Turn a 4 marker into a 2-pages-working colossus.

 

[FinalFantasy]

hi can somebody show me how to do this

integrate x^6 / sqrt(1 - x^2)

thanks

do stuff, get
I=int. sin^6@ d@
=int. sin^4@sin^2@ d@
=int. (1-cos^2@)^2sin^2@ d@
=int.(1-2cos^2@+cos^4@)sin^2@ d@
=int. sin^2 @ d@-int. 2sin^2 @cos^2 @d@+int. sin^2 @cos^4 @ d@
=1\2 int. (1-cos2@)d@-1\2int. 2sin@cos@2sin@cos@d@+int. sin^2@cos^4@d@
=1\2(@-1\2sin2@)-1\2int. sin^2 (2@)d@+int. sin^2@cos^4@d@
=1\2(@-1\2sin2@)-1\4int. (1-cos4@)d@+int. sin^2@cos^4@d@
=1\2(@-1\2sin2@)-1\4(@-1\4sin4@)+int. sin^2@cos^4@d@
=1\2(@-1\2sin2@)-1\4(@-1\4sin4@)+16(@+1\3sin^3(2@)-1\4sin(4@))


juz did it for fun, haven't checked if it's right lol.

 

[Slidey]

Yeah... so I don't see how that method is any better (shorter) than the 4 unit method, assuming you know the 4unit method.

 

[FinalFantasy]

hahaha
but this method is more "3unit like"

 

[BlackJack]

You can use (ampersand)theta(semi-colon) for easier reading... do a replace function in wordpad.

I've converted it just then, you guys should check the very last part as an exercise, i.e. S sin²θcos⁴θdθ. The rest is clean.

I=S sin⁶θ dθ
=S sin⁴θsin²θ dθ
=S (1-cos²θ )²sin²θ dθ
=S(1-2cos²θ+cos⁴θ )sin²θ dθ
=S sin² θ dθ-S 2Tin² θcos² θdθ+S sin² θcos⁴ θ dθ
=1\2 S (1-cos2θ )dθ-1\2S 2sinθcosθ2sinθcosθdθ+S sin²θcos⁴θdθ
=1\2(θ-1\2sin2θ )-1\2S sin² (2θ )dθ+S sin²θcos⁴θdθ
=1\2(θ-1\2sin2θ )-1\4S (1-cos4θ )dθ+S sin²θcos⁴θdθ
=1\2(θ-1\2sin2θ )-1\4(θ-1\4sin4θ )+S sin²θcos⁴θdθ
=1\2(θ-1\2sin2θ )-1\4(θ-1\4sin4θ )+16(θ+1\3sin³(2θ )-1\4sin(4θ ))

To nitpick: it's 1/16 and not 16.

 

[Estel]

Good 'ole reduction is best.

 

[FinalFantasy]

hahaha, thx blackjack for conversion =P
yea, oops! it's 1\16
hahaha

 

[haboozin]

I hope this helps...

ps attachment is the answer..
+ C at the end

 

[Slidey]

Integrals can be represented in different ways. However, I'm guessing that answer came from mathematica or somesuch. It's not terribly useful in terms of the HSC.

 

[haboozin]

Integrals can be represented in different ways. However, I'm guessing that answer came from mathematica or somesuch. It's not terribly useful in terms of the HSC.

sorry, i didnt mean your answer was wrong... that was just the final answer written clearly because people were complaining about reading off the long lines of working.

 

[Slidey]

Oh, no, it's no problem. But if you ever wrote that down as the answer in a test or something you'd be marked wrong. As a general rule it's the method not the answer that counts.

Unless I'm missing something and you did derive that answer...?