thanks!
Whenever you have an integral that appears to be the derivative of an inverse trig function but isn't (and cannot be solved using u-substitution), a trig substitution must be made. In your case, the integral appears to be the derivative of arcsec(x), although it is NOT. Use x = sec z.
dx = (secztanz)dz
I = int[secz*tanz*dz/(sqrt{(secz)^2-1})] = int[(sec z)(tan z)dz/sqrt((tan z)^2)] = int[secz dz] = int[{(sec z)^2 + secztanz}dz/{secz+tanz}]
where the last result was found by multiplying the top and bottom by tanz + secz.
By substitution, let u = secz+tanz. This implies the following:
du = [secztanz+(secz)^2]dz
Thus we have:
I = int[du/u] = log |u|+C = log|secz + tanz|+C = log|sec(arcsecx)+tan(arcsecx))|+ C
where log is the natural log base e and z = arcsecx.
In general, if you have a form 1/(u^2-a^2) in the integral, use u=a*secz. If you have the form 1/(u^2 + a^2) use u = a*tanz and for 1/(a^2-u^2) use u = a*sinz.
Now could somebody help me with my (much simpler) question posted in the calc threads?