integrate this (1 Viewer)

[daeneee]

what would i do to integrate:
x^2 / square root ( x^2 + 1)


im stuck on that through doing the integral of
square root ( x^2 + 1)

used integration by parts (coz its in that topic of the book) and yeah;;

help usss

 

[Iruka]

Re: integrate this slut

Let u = x; v' = x/sqrt(x^2 +1)

EDIT: you will still need to make a trig substitution to evaluate the second integral.

 

[jkwii]

Re: integrate this slut

just do this split the integral into 2 parts to get

1- 1/Sqrt(x^2-1) and then u use standard integral 1/Sqrt(x^2+1) = ln (x+Sqrt(x^2+1))

that's it

 

[independantz]

Re: integrate this slut

let u=x^2 v'=1/sqrtx^2+1

 

[Iruka]

Re: integrate this slut

Or you can try the substitution

sqrt(x^2 +1) = sin(u)

 

[malady]

I(sqrt(x^2+1)dx=x.sqrt(x^2+1)-I(x^2/sqrt(x^2+1)dx
=x.sqrt(x^2+1)-I((x^2+1-1)/sqrt(x^2+1) dx
=x.sqrt(x^2+1)-I(sqrt(x^2+1)dx+I(1/sqrt(x^2+1) dx

so I(sqrt(x^2+1)dx=1/2(x.sqrt(x^2+1) +1/2. ln(x+sqrt(x^2+1))

The orig quest then I(x^2/(sqrt(x^2+1))= 1/2(x.sqrt(x^2+1)-ln(x+sqrt(x^2+1))

 

[gypo king]

just let x=tanu

then simplify and get down to:
I:sec^3u - secu du
then integration by parts of Isec^3u and simplification give:
I:sec^3u - secu du = secutanu - I:sec^3u du + I:secu du - I:secu du
then the final answer is 0.5(secutanu)
u then draw a triangle and find cos and flip it then sub x back into tanu:

FINAL ANSWER: 0.5(xsqrt(x^2 +1))

tell me im wrong
dats alot of work though, it probably could have been done easier

 

[Affinity]

might be better: x = [e^{t} - e^{-t}]/2
then your integral =

int [e^{2t} + e^{-2t} - 2]/4 dt [the extra bits cancel.]
= e^{2t}/8 -e^{-2t}/8 - t/2
= x(sqrt(x^2+1))/2 - log(x + sqrt(x^2 + 1))/2
etc.

For those who are interested:
notice on your calculator there's a hyp button and if you press that before sthe trig functions you get sinh cosh and tanh? ever wondered what they meant?
cosh(t) = [e^{t} + e^{-t}]/2
sinh(t) = [e^{t} - e^{-t}]/2
tanh(t) = sinh(t)/cosh(t)

and you have the identities cosh^2(t) - sinh^2(t) = 1 (Used in the question)
and d/dt cosh(t) = sinh(t) and d/dt sinh(t) = cosh(t) (Used in the question)
cosh^2(t) + sinh^2(t) = cosh(2t)
2cosh^2(t) - 1 = cosh(2t) (Used in the question)
2sinh^2(t) + 1 = cosh(2t)
2cosh(t)sinh(t) = sinh(2t)

these properties make the substitution easy to compute.

usually these subsitutions work better than sec and tan subs.

 

[tacogym27101990]

回复: integrate this

yeah what the hell is these sinh(x)'s??
are we meant to know them??

 

[Slidey]

Hyperbolic sin and hyperbolic cosine.

cosh(z)=cos(iz)
2cosh(z)=e^(z)+e^(-z)

You can figure it all out using e^(ix)=cos(x)+isin(x)

 

[daeneee]

hmm yeah alot of you got it
answer is:
1/2 ( x sqrt (x^2 + 1) + log( x + sqrt (x^2 + 1)) )

cheers for the workin out guys
much appreciated~



*EDIT* that answer above is the answer to the WHOLE quesiton, and not of the part that i initially askedd

 

[Affinity]

hmm yeah alot of you got it
answer is:
1/2 ( x sqrt (x^2 + 1) + log( x + sqrt (x^2 + 1)) )

cheers for the workin out guys
much appreciated~



*EDIT* that answer above is the answer to the WHOLE quesiton, and not of the part that i initially askedd

1/2 ( x sqrt (x^2 + 1) - log( x + sqrt (x^2 + 1)) )