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Thread: integration by parts

  1. #1
    Junior Member
    Jan 2009

    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
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  2. #2
    Super Member redsoxfan325's Avatar
    Feb 2009
    Swampscott, MA
    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, $\displaystyle du=\frac{1}{t\sqrt{t^2-1}}dt$ and $\displaystyle v=\frac{t^2}{2}$. Thus, your integral becomes $\displaystyle \frac{t^2}{2}sec^{-1}(t)-\frac{1}{2}\int \frac{t^2}{2t\sqrt{t^2-1}}\,dt$ = $\displaystyle \frac{t^2}{2}sec^{-1}(t)-\int \frac{t}{2\sqrt{t^2-1}}\,dt$. Now you can do a u-substitution. Let $\displaystyle u=t^2-1$ so that $\displaystyle du=2tdt$. Subbing in leaves you with $\displaystyle \frac{t^2}{2}sec^{-1}(t)-\frac{1}{2}\int \frac{1}{2\sqrt{u}}\,du = \frac{t^2}{2}sec^{-1}(t)-\frac{1}{2}\sqrt{u} = \frac{t^2}{2}sec^{-1}(t)-\frac{1}{2}\sqrt{t^2-1}$. Thus the answer is $\displaystyle \frac{t^2}{2}sec^{-1}(t)-\frac{1}{2}\sqrt{t^2-1} + C$.

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