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Integration by substitution help

  1. Mar 1, 2005 #1
    I am going crazy on this problem:

    [tex] \int sec(v+(\pi/2)) tan(v+\pi/2)) dv [/tex]

    if I substitute u= [tex] tan(v+\pi/2)) dv [/tex], can I use the product rule to find du= [tex] sec(v+(\pi/2)) dv [/tex].

    Thanks, Todd
     
  2. jcsd
  3. Mar 1, 2005 #2

    Galileo

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    Use [itex]\sin(x+\pi/2)=\cos(x)[/itex] and [itex]\cos(x+\pi/2)=-\sin(x)[/itex] to rewrite the integrand. Then subsitute [itex]u=\frac{1}{\sin(x)}[/itex].
     
  4. Mar 1, 2005 #3

    dextercioby

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    Else,use the definition and the substitution [tex] x+\frac{\pi}{2}=u [/tex]...It's really simple.

    And another one:
    [tex]d[\sec(x+\frac{\pi}{2})]=\sec(x+\frac{\pi}{2})\tan(x+\frac{\pi}{2})dx [/tex]

    so the integration is immediate...

    Daniel.
     
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