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Evaluating Trig' Integrals

  1. Aug 7, 2005 #1
    Can someone please help me in evaluating the following integral?

    Im really bad with trig functions, and have been trying to work this out for ages.

    [tex]\int \frac{sin x}{cos^2 x} dx[/tex]

    Is there a way I can evaluate for [tex]\cos^2 x[/tex] ??????
    Or is there something Im missing altogether?


    Can someone please tell me how:
    [tex]\int \frac{sec\theta}{cos\theta} dx =\ tan\theta + C[/tex]
    Last edited: Aug 7, 2005
  2. jcsd
  3. Aug 7, 2005 #2


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    Substitution: [itex]y = \cos x \Leftrightarrow dy = - \sin xdx[/itex] gives:

    [tex]\int {\frac{{\sin x}}
    {{\cos ^2 x}}dx} = \int {\frac{{ - 1}}
    {{y^2 }}dy = } \frac{1}
    {y} + C = \frac{1}
    {{\cos x}} + C[/tex]

    If you know that :
    [tex]\sec \theta = \frac{1}{{\cos \theta }}[/tex]


    [tex]\frac{{d\left( {\tan \theta } \right)}}
    {{d\theta }} = \frac{1}
    {{\cos ^2 \theta }}[/tex]

    then... :smile:
  4. Aug 8, 2005 #3

    For the first integral [tex]\int \frac{sin x}{cos^2 x} dx[/tex] was thinking more along the lines of multiplying out before evaluatng, to try to obtain a more simple integral to evaluate,
    but your way is obviously better.

    Is there a way this can be evaluated without substitution?
    Or is there no function which has [tex]\cos^2 x[/tex] as its derivative?

    Thanks again for the help.
    Last edited: Aug 8, 2005
  5. Aug 8, 2005 #4


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    why would you want cos^2@ as a derivative?
  6. Aug 8, 2005 #5


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    The easiest way is certainly using that substitution.

    I don't really see why you'd want to know an anti-derivative for [tex]\cos^2 x[/tex], if you do: just compute it, use:
    [tex]\cos \left( {2x} \right) = 2\cos ^2 x - 1 \Leftrightarrow \cos ^2 x = \frac{{1 + \cos \left( {2x} \right)}}

    I don't see how that would help though, you don't have [tex]\cos^2 x[/tex] in the integral but [tex]\frac{1}
    {{\cos ^2 x}}[/tex], whose anti-derivative is of course [tex]\tan x[/tex]
  7. Aug 8, 2005 #6
    [tex]\int \frac{sin x}{cos^2 x} dx = \int \sec x \tan x dx = \sec x + C [/tex]
  8. Aug 9, 2005 #7
    Thanks very much for all your help everyone.

    Im slowly getting the hang of Calculus, but im not at all good with derivatives of trig functions.
    Ill get there eventually.
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