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Math Help - direct integration

  1. #1
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    direct integration

    Hello MHF, please help me with this problem

    \int tan(x/2).sec^2(x/2)+sec^2(x/2)
    t=(x/2)
    \int tan(t). sec^2(t)dt+tan(t) = -ln[cos(t)]+tan(t)+C

    But isnt correct, i'm afraid my substitution is not correct
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  2. #2
    MHF Contributor MarkFL's Avatar
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    Re: direct integration

    If your original integral is:

    \int \tan\left(\frac{x}{2} \right)\sec^2\left(\frac{x}{2} \right)+\sec^2\left(\frac{x}{2} \right)\,dx

    I would first factor the integrand:

    \int \sec^2\left(\frac{x}{2} \right)\left(\tan\left(\frac{x}{2} \right)+1 \right)\,dx

    Now, let:

    u=\tan\left(\frac{x}{2} \right)+1\,\therefore\,du=\frac{1}{2}\sec^2\left( \frac{x}{2} \right)\,dx

    and the integral becomes:

    2\int u\,du

    Now, find the anti-derivative, then back-substitute for u.
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  3. #3
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    Re: direct integration

    Quote Originally Posted by MarkFL2 View Post
    If your original integral is:

    \int \tan\left(\frac{x}{2} \right)\sec^2\left(\frac{x}{2} \right)+\sec^2\left(\frac{x}{2} \right)\,dx

    I would first factor the integrand:

    \int \sec^2\left(\frac{x}{2} \right)\left(\tan\left(\frac{x}{2} \right)+1 \right)\,dx

    Now, let:

    u=\tan\left(\frac{x}{2} \right)+1\,\therefore\,du=\frac{1}{2}\sec^2\left( \frac{x}{2} \right)\,dx

    and the integral becomes:

    2\int u\,du

    Now, find the anti-derivative, then back-substitute for u.
    MarkFL i'm learning now integral in school and we dont use substitution yet(integration techniques)
    .
    Last edited by Chipset3600; November 13th 2012 at 03:46 PM.
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  4. #4
    MHF Contributor MarkFL's Avatar
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    Re: direct integration

    Quote Originally Posted by Chipset3600 View Post
    ...i'm afraid my substitution is not correct
    Quote Originally Posted by Chipset3600 View Post
    ...we dont use substitution yet...
    Do you see why I assumed you wanted to use a substitution?
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  5. #5
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    Re: direct integration

    Quote Originally Posted by MarkFL2 View Post
    Do you see why I assumed you wanted to use a substitution?
    Yes, sorry! I did it just to ease my calculations
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  6. #6
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    Re: direct integration

    This is a difficult integral for someone who hasn't learned about u-substitution yet. You're probably aware that the derivative of \tan{\frac{x}{2}} is \frac{1}{2}\sec^2{\frac{x}{2}}. And looking at the other term might eventually lead you to take the derivative of \tan^2{\frac{x}{2}}. Once you do that, you should be able to combine the two derivatives to get the expression you want to integrate.

    It's probably less work to first learn u-substution, then apply it to this integral. But I guess that's not an option for you.

    - Hollywood
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  7. #7
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    Re: direct integration

    Quote Originally Posted by hollywood View Post
    This is a difficult integral for someone who hasn't learned about u-substitution yet. You're probably aware that the derivative of \tan{\frac{x}{2}} is \frac{1}{2}\sec^2{\frac{x}{2}}. And looking at the other term might eventually lead you to take the derivative of \tan^2{\frac{x}{2}}. Once you do that, you should be able to combine the two derivatives to get the expression you want to integrate.

    It's probably less work to first learn u-substution, then apply it to this integral. But I guess that's not an option for you.

    - Hollywood
    Actually isnt so hard, take a look:

    \int tan(x/2)sec^2(x/2)dx.(\frac{1}{2}.\frac{2}{1})  + \int sec^2(x/2)dx.(\frac{1}{2}.\frac{2}{1})

    2\int tan(x/2).\frac{1}{2} sec^2(x/2)dx + 2\int sec^2(x/2). \frac{1}{2}dx

    Now i use the power rule for the first and the sec^2(u) rule for the second integral

    tan^2(x/2)+ 2tan(x/2)+C
    Now take a look of the derivative of my answer: direct integration-derivada.gif
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