Integral Help - $\int \frac {\sqrt{u^2-a^2}}{u}du$
As per the integral table, how do I show $\int \frac {\sqrt{u^2-a^2}}{u}du$ = $ {\sqrt{u^2-a^2}} - \arccos^{-1}(\frac {a}{u})+c$
I can't figure out how to start it.
I can do it for: $$ \int \frac {\sqrt{u^2-a^2}}{u}du = \sqrt{u^2-a^2} - \mathrm{arcsec}^{-1}(\frac{u}{a})+c $$ by starting with $\mathrm{arcsec}\ \theta = u$
Is it just a matter of showing the relationship between $\mathrm{arcsec}$ and $\arccos$?
Yep. Since $\sec x=1/\cos x$, if we set $$\sec x=1/\cos x=y,$$ then $$x=\sec^{-1}y$$ and $$x=\cos^{-1}\frac1y,$$ so $\sec^{-1}y=\cos^{-1}\frac1y$.