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Simple Integration

  1. Jan 18, 2008 #1
    [SOLVED] Simple Integration

    1. Integrate the following

    f(x) = sin x/(cos x )^2

    3. Well i dont really want to use subsititution or anything since im pretty sure this can be done very simply but i dont know why i cannot get it...

    sin x /cos ^2 x = sinx / 1- sin^2x

    i know ln (cos x)^2 will give me (-2(cosx)*sinx)(1/cos^2x)

    so thats not good...
  2. jcsd
  3. Jan 18, 2008 #2
    How about u-substitution?
  4. Jan 18, 2008 #3
    You must use substitution. Using trig identities will only make it messier.

    Hint: get f(x) into the form du / u ^2 .
  5. Jan 18, 2008 #4
    yea i just did it with u as well but that weird, cause this question was in my first section for integral calc (where we basically onyl went through basic rules like power rule, and integrals of other trig functions)

    but anyways

    let u = cosx

    du/dx = -sinx

    now i integrate - 1/(u)^2 * du
    = integral of -u^-2 *du

    = u^-1

    = cos^-1 x

    = 1/cos x

    = sec x

  6. Jan 18, 2008 #5


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    Homework Helper

    Yup, that seems perfectly correct. Except for a small error, you forgot the Constant of Integration "+ C" at the end of the final result. :)
  7. Jan 18, 2008 #6


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    Homework Helper

    It happens in this case that there is another way to go about this, although it's not something you'd generally spot at first (you notice it after doing the u-substitution). You can also write
    (sin x)/[(cos x)^2] as (1/cos x)·(sin x / cos x) = sec x tan x , which is the derivative of sec x . (This at least serves as a check on your result...)
  8. Jan 20, 2008 #7
    [tex]\int\frac{\sin x}{\cos^2 x}dx=-\int\frac{1}{\cos^2 x}d(\cos x)=-\int (\cos x)^{-2}d(\cos x)=-\frac{(\cos x)^{-2+1}}{-2+1}+C=\frac{1}{\cos x} +C[/tex]
    [tex]d(\cos x)=-\sin x dx[/tex]
    [tex]dx=\frac{d(\cos x)}{-\sin x}[/tex]
    Last edited: Jan 20, 2008
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