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  1. #1
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    integral

    integrate dx/tanxcos^2x

    sinx=u and du=cosxdx and dxcosx/sinxcos^2x then du/u(1-u)^2
    1/u+1/1-u+1/1+u
    a/u + b/1-u + c/1+u
    I found c=-1/2, a=1 and b=1/2

    ln|u|+1/2ln|1-u|-1/2ln|1+u|

    ln|sinx|+1/2ln|1-sinx|-1/2ln|1+sinx|
    ln|sinx|+ln|1-sinx|/1/2ln|1+sinx|

    and I got stuck here.How can I solve that?
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  2. #2
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    Re: integral

    integral-11-jun-13.png
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  3. #3
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    Re: integral

    Iwant to solve it in my way.
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  4. #4
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    Re: integral

    Quote Originally Posted by kastamonu View Post
    Iwant to solve it in my way.
    Then why did you ask for help?

    Also, why don't you learn to post in a readable format.

    Here is what you actually posted \int {\frac{{dx}}{{\tan (x)}}{{\cos }^2}(x)} .

    But it appears that is not what you mean, If so, learn to use grouping symbols.

    Now note that \int {\frac{{dx}}{{\tan (x)}\cos^2(x)} =\int {\frac{{\sec^2(x)dx}}{{\tan (x)}}

    If you do not recognize at once that as a logarithm, then you are crippled by u-substitutions.
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  5. #5
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    Re: integral

    This is very difficult to read! Please use lots of parentheses to make your meaning clear.
    For example, "b/1- u" would, strictly, be read "b- u" but I am sure you mean b/(1- u).

    Quote Originally Posted by kastamonu View Post
    integrate dx/tanxcos^2x

    sinx=u and du=cosxdx and dxcosx/sinxcos^2x then du/u(1-u)^2
    1/u+1/1-u+1/1+u
    a/u + b/1-u + c/1+u
    I found c=-1/2, a=1 and b=1/2

    ln|u|+1/2ln|1-u|-1/2ln|1+u|

    ln|sinx|+1/2ln|1-sinx|-1/2ln|1+sinx|
    ln|sinx|+ln|1-sinx|/1/2ln|1+sinx|
    A typo. You mean ln|sin(x)|+ (1/2)ln(|1- sin(x)|)- (1/2)ln(|1+ sin(x)|)

    and I got stuck here.How can I solve that?
    What do you mean by "solve" it? You were asked to find an integral and you have. I presume you mean "simplify it" or write it in a simpler form. To do that use the laws of logarithms:
    cln(a)= ln(a^c) so (1/2)ln(|1- sin(x)|)= ln(|1- sin(x)|^{1/2})= ln(\sqrt{1- sin(x)}
    and (1/2)ln(|1+ sin(x)|)= ln(|1+ sin(x)|^{1/2}= ln(\sqrt{1+ sin(x)})

    ln(a)- ln(b)= ln(\frac{a}{b}) so
    (1/2)ln(|1- sin(x)|)- (1/2)ln(|1+ sin(x)|)= ln(\sqrt{1- sin(x)})- ln(\sqrt{1+ sin(x)})= ln\left(\frac{\sqrt{1- sin(x)}}{\sqrt{1+ sin(x)}}\right)= ln\left(\sqrt{\frac{1- sin(x)}{1+ sin(x)}}\right)

    ln(a)+ ln(b)= ln(ab) so
    ln|sin(x)|+ (1/2)ln(1- sin(x))- (1/2)ln(1+ sin(x))= ln|sin(x)|ln\left(\sqrt{\frac{1- sin(x)}{1+ sin(x)}}\right)= ln\left|sin(x)\sqrt{\frac{1- sin(x)}{1+ sin(x)}}\right|
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    Re: integral

    I was stuck. By this reason I asked for your help. I don't like memorizing formulas so I wanted to solve it in that way. I made a mistake in "1/1-u".I found the answer.Many Thanks.
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  7. #7
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    Re: integral

    Quote Originally Posted by Plato View Post
    Then why did you ask for help?

    Also, why don't you learn to post in a readable format.

    Here is what you actually posted \int {\frac{{dx}}{{\tan (x)}}{{\cos }^2}(x)} .

    But it appears that is not what you mean, If so, learn to use grouping symbols.

    Now note that \int {\frac{{dx}}{{\tan (x)}\cos^2(x)} =\int {\frac{{\sec^2(x)dx}}{{\tan (x)}}

    If you do not recognize at once that as a logarithm, then you are crippled by u-substitutions.
    I made a mistake in "1/1-u".I found the answer.Thanks.
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  8. #8
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    Re: integral

    Quote Originally Posted by kastamonu View Post
    I was stuck. By this reason I asked for your help. I don't like memorizing formulas so I wanted to solve it in that way. I made a mistake in "1/1-u".I found the answer.Many Thanks.
    I agree with you that it is pointless to remember too many formulas, as they simply get mixed up and end up confusing people. However, \displaystyle \frac{d}{dx} \left[ \tan{(x)} \right] = \sec^2{(x)} is a standard derivative, so is one that I and you should memorise.
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