1. ## integrate

$\displaystyle \int {\frac{{dx}} {{\sqrt {x^2 - 1} }}}$

thanks!

2. try using $\displaystyle x = \sec(\theta)$

3. I dont know how to derive it but if my memory serves me right.

$\displaystyle \int {\frac{{dx}} {{\sqrt {x^2 - 1} }}} = \cosh^{-1}(x)+C$

Just try taking the derivative of $\displaystyle \cosh(x)$

4. ## solution brah

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?