Trig sine substitution doesnt work

[stunner5000pt]

how would on integrate \int_{0}^{1} \sqrt{1 + 4t^2} dt
trig sub sittution doesn't work since one doesn't 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]

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

Regards,
George

 

[CarlB]

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]

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]

Lots of ways to skin a small furry critter.
Carl

AbsolutelY!

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

Regards,
George

 

[stunner5000pt]

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