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Math Help - how can solve this integral

  1. #1
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    how can solve this integral

    hi!


    How can I start with this integral

    tanx/cos x^2 dx

    thanks
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  2. #2
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    A couple options. I will do one to completion (a way that I suspect they do not want you to do), and will get you started on the way I'd imagine they want you to do this:

    1st Method -

    \int \frac{tan(x)dx}{cos^{2}(x)}\Rightarrow

    =\int \frac{sin(x)dx}{cos^{3}(x)}

    Make the substitution:

    u=cos(x); -du=sin(x)dx

    =\int \frac{sin(x)dx}{cos^{3}(x)} \Rightarrow -\int \frac{du}{u^{3}}

    =\frac{1}2{u^2}

    =\frac{1}{2cos^{2}(x)}

    The second method begins much in the same way by rewritting the integral:

    \int \frac{tan(x)dx}{cos^{2}(x)}\Rightarrow

    \int \frac{tan(x)sec(x)dx}{cos(x)}

    See if you can take it from there.
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  3. #3
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    I would think that

    \int \frac{\tan x}{\cos^2x}dx=\int \tan x\cdot\sec^2xdx

    And if u=\tan x, then du=\sec^2xdx and...

    \int u\cdot du=\frac{u^2}{2}+C=\frac{\tan^2x}{2}+C
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  4. #4
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    Quote Originally Posted by hoger View Post
    hi!


    How can I start with this integral

    tanx/cos x^2 dx

    thanks
    Is it \int{\frac{\tan{x}}{\cos^2{x}}\,dx} or \int{\frac{\tan{x}}{\cos{(x^2)}}\,dx}?
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