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Homework Help: Difficult integral

  1. Mar 20, 2007 #1
    1. The problem statement, all variables and given/known data
    [tex] \int \frac{\sin x+\cos x}{\sec x+ \tan x}dx [/tex]

    2. Relevant equations

    [tex] \sin x = \frac{1}{\sec x} [/tex]
    [tex] \cos x = \frac{\sin x}{\tan x} [/tex]

    3. The attempt at a solution
    i separate and try to use identities but i got nothign

    1/(secx^2+secx tan x)+sin x/tanx^2+sec x :confused:
    Last edited: Mar 20, 2007
  2. jcsd
  3. Mar 20, 2007 #2


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    Homework Helper

    write everything in terms of sin x and cos x only, then integrate with change of variable.
  4. Mar 20, 2007 #3


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    Staff Emeritus
    Science Advisor

    One of your identities is incorrect: sec(x)=1/cos(x).

    As has been said above, you should first look to express everything in terms of sines and cosines. See if this gives you a hint as to how to proceed.
  5. Mar 21, 2007 #4
    please tell more

    this is the shape that i got
    don't know how to complete

    –integral sin²x-sin2x-1/2(1+sinx)
  6. Mar 22, 2007 #5

    Gib Z

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    Homework Helper

    [tex]-(\int \sin^2 x dx - \int \sin 2x dx - 1/2\int 1+\sin x)[/tex]

    Write sin^2 x as (1/2) (1-cos2x).

    for sin 2x, make a substitution u=2x, then remember the integral of sin u is -cos u.

    For the 3rd one, if you can't do it, why are you doing this question?
  7. Mar 24, 2007 #6
    you can also try weierstrass substitution
    i.e t = tan x/2
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