
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

I'm not entirely sure what you are trying to integrate.
If you're trying to integrate $\displaystyle \int \frac{t}{sec(x)}\,dt$, that just equals $\displaystyle \int tcost(t)\,dt$. At this point, let $\displaystyle u=t$ and $\displaystyle dv=cos(t)dt$. Thus $\displaystyle du=dt$ and $\displaystyle v=sin(t)$. Thus your integral becomes $\displaystyle tsin(t)\int sin(t)\,dt$. Thus the answer is $\displaystyle tsin(t)+cos(t) + C$.
If you're trying to integrate $\displaystyle \int tsec^{1}(t)\,dt$, let $\displaystyle u=sec^{1}(t)$ and $\displaystyle dv=tdt$. Thus, $\displaystyle du=\frac{1}{t\sqrt{t^21}}dt$ and $\displaystyle v=\frac{t^2}{2}$. Thus, your integral becomes $\displaystyle \frac{t^2}{2}sec^{1}(t)\frac{1}{2}\int \frac{t^2}{2t\sqrt{t^21}}\,dt$ = $\displaystyle \frac{t^2}{2}sec^{1}(t)\int \frac{t}{2\sqrt{t^21}}\,dt$. Now you can do a usubstitution. Let $\displaystyle u=t^21$ so that $\displaystyle du=2tdt$. Subbing in leaves you with $\displaystyle \frac{t^2}{2}sec^{1}(t)\frac{1}{2}\int \frac{1}{2\sqrt{u}}\,du = \frac{t^2}{2}sec^{1}(t)\frac{1}{2}\sqrt{u} = \frac{t^2}{2}sec^{1}(t)\frac{1}{2}\sqrt{t^21}$. Thus the answer is $\displaystyle \frac{t^2}{2}sec^{1}(t)\frac{1}{2}\sqrt{t^21} + C$.
I hope that one of these was what you're looking for.