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Math Help - Trig Integral help

  1. #1
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    Trig Integral help

    Indefinite integral of dx/(sin(x)+tan(x)

    I replaced tan(x) with sin(x)/cos(x) so then I flipped the sinx/cosx term and wrote the integral as:

    cosx/(sin(x) + sin(x) dx

    becomes

    cosx/(2 sinx) dx

    Then I took the 1/2 out of the integral

    1/2 times integral cos x / sin x dx

    Let u = sin x du = cos x dx

    1/2 times integral du / u

    integrate

    1/2 * ln u + C

    replace sin x

    1/2 * ln sin x + C

    Is that correct?

    And how do you do this with replacing sin x with 2u/ 1 + u^2 in the original problem and so on?

    kind of hung up that way with the tangent function?

    Thanks
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  2. #2
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    Quote Originally Posted by PHJ426 View Post
    1/2 * ln sin x + C

    Is that correct?
    You should have absolute values.
    And remember to check all you need to do is take the derivative and see if it works out!
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  3. #3
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    Quote Originally Posted by PHJ426 View Post
    Indefinite integral of dx/(sin(x)+tan(x)

    I replaced tan(x) with sin(x)/cos(x) so then I flipped the sinx/cosx term and wrote the integral as:

    cosx/(sin(x) + sin(x) dx
    There is a ")" missing in your first integral. Is this \int \frac{dx}{\sin(x)+ \tan(x)}?

    If so then \frac{1}{\sin(x)+ \tan(x)} = \frac{1}{\sin(x)+ \frac{\sin(x)}{\cos(x)}} and multiplying both numerator and denominator by cos(x) gives \frac{\cos(x)}{\sin(x) \cos(x)+ \sin(x)}, NOT what you have.

    becomes

    cosx/(2 sinx) dx

    Then I took the 1/2 out of the integral

    1/2 times integral cos x / sin x dx

    Let u = sin x du = cos x dx

    1/2 times integral du / u

    integrate

    1/2 * ln u + C

    replace sin x

    1/2 * ln sin x + C

    Is that correct?

    And how do you do this with replacing sin x with 2u/ 1 + u^2 in the original problem and so on?

    kind of hung up that way with the tangent function?

    Thanks
    Last edited by mr fantastic; February 27th 2009 at 03:53 AM. Reason: Fixed some latex
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