# Trig sine substitution doesnt work

1. Jan 12, 2006

### stunner5000pt

how would on integrate $$\int_{0}^{1} \sqrt{1 + 4t^2} dt$$
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.

2. Jan 12, 2006

### George Jones

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

Regards,
George

3. Jan 12, 2006

### 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

4. Jan 12, 2006

### 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.

5. Jan 12, 2006

### George Jones

Staff Emeritus
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

6. Jan 12, 2006

### stunner5000pt

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