Thread: Intergrating function to power of 1/2

I'm having trouble integrating this arc length.

General formula:

, between intervals a and b

My function: , where a=0.00183

So

Putting this derivative into general formula:

, between the intervals of 176.3 and -176.3. (I haven't put a=0.00183 in yet because it would get too messy intergrating it.)

I don't know how to integrate this because I think it involves some method of substitution which I don't know. Any help would be great ;D

If CAS are allowed:

Function calculator

Whenever you have 1+x^2 or 1-x^2 under the square root in an integral it would be a good idea to try some sort of trig substiution using either
sin^2 +cos^2=1
or
cosh^2-sinh^2=1

in this case of course you would use the hyperbolic identity.
express 2ax=cosh(t)
calculate the derivative and make the substitution.

you should end up with
(1/2a) int (sinh^2)

express that in terms of e^x and its easy enough to integrate.