Trig sine substitution doesnt work

how would on integrate [tex] \int_{0}^{1} \sqrt{1 + 4t^2} dt [/tex]
trig sub sittution doesnt work since one doesnt get tan^2 +1 .

i tryed solving this with matematica and it yielded something with a sinh argument. I am not familiarwi the hyp sine substitution.
 

George Jones

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Use the sustitution [itex]2t = tan \theta[/itex]. Be prepared to integrate [itex]sec^3 \theta[/itex].

Regards,
George
 

CarlB

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You may find that a substitution of t for sinh(theta)/2 will assist. And then you might find it useful to know how to write cosh(theta) in terms of exponentials.

Lots of ways to skin a small furry critter.

Carl
 

GCT

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to make things appear simpler factor out the 4, then make the proper trig substitution; if you still don't get it, you've definitely got some more studying to do.
 

George Jones

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CarlB said:
Lots of ways to skin a small furry critter.
Carl
AbsolutelY!

Another way is to make the substitution [itex]2t = sinh x[/itex], and then to make a "half-angle" substitution for [itex]cosh^{2} x[/itex].

Regards,
George
 
would the limits of integration change in this process?
for t=0, theta = 0
for t =1 , theta = arctan 2
is that right?
 

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