Quickest way to integrate (1 Viewer)

[ezzy85]

whats the best way to integrate something like this:

x sqrt(x² - 8x) or just sqrt(x² - 8x)

id prefer not to do it by parts.
thanks

 

[Affinity]

Hmm probably repeating what you know, I don't know better methods

I sqrt(x^2 - 8x) dx
= I sqrt[ (x-4)^2 - 16 ] dx

since you don't want to do it by parts we have hyperbolic substitution,

definitions cosh(t)= [e^x + e^(-x)]/2 , sinh(t)= [e^x - e^(-x)]/2

let (x-4) = 4cosh(t), dx = 4sinh(t) dt

above integral
= I sqrt( 16cosh(t)^2 - 16 ) * 4sinh(t) dt
= I sqrt( 16sinh(t)^2 ) * 4sinh(t) dt
= I 16sinh(t)^2 dt
= 8 I [cosh(2t) - 1] dt
= 8[ (sin(2t)/2) - t] + C
= 8[ sinh(t)cosh(t) - t] + C

now because (x-4) = 4cosh(t)
sinh(t) = sqrt( cosh(t)^2 - 1) = sqrt( ((x-4)^2 / 16) - 1)
t = arccosh[(x-4)/4] = ln { (x - 4)/4 + sqrt[ (x-4)^2 / 16 - 1] }

so the original integral
= 8 * [ (1/4)(x-4)sqrt( ((x-4)^2 / 16) - 1) +ln { (x - 4)/4 + sqrt[ (x-4)^2 / 16 - 1] } ] + C

OR

the usual sec subsitution:

let x-4 = 4sec(t) , then dx = 4sec(t)tan(t) dt

= I sqrt[ 16sec(t)^2 - 16 ] * 4sec(t)tan(t) dt
= I 4tan(t) * 4sec(t)tan(t) dt
= I 16sec(t)tan(t)^2 dt
hmm still by parts.. which is not good..
if K = I sec(t)tan(t)^2 dt
then
K = sec(t)tan(t) - I sec(t)^3 dx
K = sec(t)tant(t) - I sec(t)tan(t)^2 dx - I sec(t) dx
K = sec(t)tant(t) - K - ln {sec(x) + tan(x)} + C
K = 0.5 *[ sec(t)tan(t) - ln{ sec(x) + tan(x) } ] + C
so the original integral
= 8 * [ sec(t)tan(t) - ln{ sec(x) + tan(x) } ]+ C
by (x-4) = 4sec(t), tan(t) = sqrt[ (x-4)^2 / 16 - 1]
substitute,
= 8 * [ (1/4)(x-4)sqrt[ (x-4)^2 / 16 - 1] - ln { (x - 4)/4 + sqrt[ (x-4)^2 / 16 - 1] } ] + C

Hmm.. about the same..

 

[ezzy85]

ill stick with the sec substitution. thanks for that.

 

[deyveed]

you know how you use cosh, arccosh, arctan etc
what do the 'arc' and 'h' bits do?

 

[ezzy85]

arc is inverse and im not sure about h, which is why im sticking with sec.

 

[Lazarus]

The 'h' stands for hyperbolic (see Affinity's definitions).

 

[deyveed]

What is a hyperbolic substitution?

 

[Affinity]

it's just like how you do a x=sin(t) substitution to integrate something like

1/sqrt(1-x^2)

with hyperbolic substitutions you do a
x=cosh(t)=[e^(t) + e^(-t)]/2 subsitution
sinh(t)=[e^(t) - e^(-t)]/2

other hyperbolic functions such as tanh could be defined accordingly

some properties include

[cosh(t)]^2 - [sinh(t)]^2 = 1

cosh(a+b)=cosh(a)cosh(b) + sinh(a)sinh(b)

sinh(a+b)=sinh(a)cosh(b) + cosh(a)sinh(b)

cosh(2c)=[cosh(c)]^2 + [sinh(c)]^2

sinh(2c)=2cosh(c)sin(c)

d/dt [ cosh(t) ] = sinh(t)

d/dt [ sinh(t) ] = cosh(t)

geometrically, cosh(t) sinh(t) lies on the right arm of the unit rectangular hyperbola, and the area bounded by the hyperbola, the x axis and the ray from the origin to that point is t/2.

 

[Rahul]

excuse my stupidity, but i have never come across the term 'cosh'. would you be able to direct me to somewhere in a text or explain here if not too much of a problem.

 

[freaking_out]

Originally posted by Rahul
excuse my stupidity, but i have never come across the term 'cosh'. would you be able to direct me to somewhere in a text or explain here if not too much of a problem.

its not in ze syllabus...so u prolly betta off lookin' at uni textbooks.

 

[Wacky]

There was also an article in parabola way back telling you how to use Sinh and Cosh functions to solve cubics. It was really cool. but it isn't in the syllabus so forget about it cos markers = stupid, allegedly.

 

[Affinity]

but I thought you only need to find the correct answer for integrals

 

[freaking_out]

Originally posted by Affinity
but I thought you only need to find the correct answer for integrals

i heard from terry lee (the author & senior marker), that this guy used hyberbolic substitutions to do a question, and at the end they had to give him no marks, coz the method is beyond the syllabus...so i don't know...maybe u should also ask the hsc advice line.