Intergra. (1 Viewer)

[TheCanuck]

Guys...how do u do these questions...Im sort of stuck here....

intergral of dx/94-x^2)

intergral of dx/(x^2-4)

thanx guys =)

 

[drynxz]

hmm ..theres use to be a standard intergral for this..but i dont think they teach it anymore.

i think its

int. dx/(a^2 - x^2) = 1/2a ln [(a+x)/(a-x)] + c

and

int. dx/(x^2-a^2) = 1/2a ln [(x-a)/(x+a)] + c

 

[jb_nc]

hmm ..theres use to be a standard intergral for this..but i dont think they teach it anymore.

i think its

int. dx/(a^2 - x^2) = 1/2a ln [(a+x)(a-x)] + c

and

int. dx/(x^2-a^2) = 1/2a ln [(x-a)/(x+a)] + c

check this: http://www.maths.usyd.edu.au/u/UG/JM/MATH1605/r/assign/standard_integrals.pdf

 

[Shortbreads]

You could always try substitutions like x= sqrt(94)sin@ for the first one and x= 2sec@ for the second.

Or take out a minus and use x= 2sin@, which might be easier.

 

[nickthebiscuit]

hmm ..theres use to be a standard intergral for this..but i dont think they teach it anymore.

i think its

int. dx/(a^2 - x^2) = 1/2a ln [(a+x)(a-x)] + c

and

int. dx/(x^2-a^2) = 1/2a ln [(x-a)/(x+a)] + c

int. dx/(a^2 - x^2) = 1/2a ln [(a+x)/(a-x)] + c

yeah you missed the divide sign

 

[SoulSearcher]

Guys...how do u do these questions...Im sort of stuck here....

intergral of dx/94-x^2)

intergral of dx/(x^2-4)

thanx guys =)

You can always use partial fractions, as a last resort.

 

[KFunk]

Substitute x=2sin@ into the first one, so that dx = 2cos@d@, yielding

(Int) (2cos@)d@/(4 - 4sin²@)

= 1/2 (Int) Sec@ d@

You then have a painful process: http://www.math.com/tables/integrals/more/sec.htm

= 1/2 ln|sec@ + tan@| + C

---------- Converting back to x ---------------

Make a right angle triangle where sin@ = x/2, giving sec@ = 2/Sqrt(4-x²) and tan@ = x/Sqrt(4-x²). Hence we obtain:

1/2 ln|(x+2)/Sqrt(4-x²)| + C

= 1/2 ln|Sqrt[(x+2)²/(2+x)(2-x)]| + C

= 1/4 ln|(2+x)/(2-x)| + C ..... by moving sqrt out of log as 1/2 and cancelling (x+2)

 

[toadstooltown]

Eurgh. I never understood why people substitute. They can both be simply done by partial fractions and easily gets you to the correct logs.

 

[Shortbreads]

I'd like to take that wink in a suggestive manner...

But it's pretty much impossible in this case.

*sigh*

 

[drynxz]

int. dx/(a^2 - x^2) = 1/2a ln [(a+x)/(a-x)] + c

yeah you missed the divide sign

woops its fixed now