1. ## integration by parts

how do i find the integral of t*sec(t)^-1 dt ?
i figure set u=t v=ln(sec(t))
du=1 dt dv=sec(t)^-1 dt
I'm going to use S for the integral symbol.

then i'd get t*ln(sec(t)) - Sln(sec(t))

is this correct so far? how do i find the integral of ln(sec(t))? i don't know what to do at this point

2. I'm not entirely sure what you are trying to integrate.

If you're trying to integrate $\int \frac{t}{sec(x)}\,dt$, that just equals $\int tcost(t)\,dt$. At this point, let $u=t$ and $dv=cos(t)dt$. Thus $du=dt$ and $v=sin(t)$. Thus your integral becomes $tsin(t)-\int sin(t)\,dt$. Thus the answer is $tsin(t)+cos(t) + C$.

If you're trying to integrate $\int tsec^{-1}(t)\,dt$, let $u=sec^{-1}(t)$ and $dv=tdt$. Thus, $du=\frac{1}{t\sqrt{t^2-1}}dt$ and $v=\frac{t^2}{2}$. Thus, your integral becomes $\frac{t^2}{2}sec^{-1}(t)-\frac{1}{2}\int \frac{t^2}{2t\sqrt{t^2-1}}\,dt$ = $\frac{t^2}{2}sec^{-1}(t)-\int \frac{t}{2\sqrt{t^2-1}}\,dt$. Now you can do a u-substitution. Let $u=t^2-1$ so that $du=2tdt$. Subbing in leaves you with $\frac{t^2}{2}sec^{-1}(t)-\frac{1}{2}\int \frac{1}{2\sqrt{u}}\,du = \frac{t^2}{2}sec^{-1}(t)-\frac{1}{2}\sqrt{u} = \frac{t^2}{2}sec^{-1}(t)-\frac{1}{2}\sqrt{t^2-1}$. Thus the answer is $\frac{t^2}{2}sec^{-1}(t)-\frac{1}{2}\sqrt{t^2-1} + C$.

I hope that one of these was what you're looking for.