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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 \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.
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