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Math Help - [SOLVED] how to start integrating e^(-x)*tan(e^(-x))dx

  1. #1
    Newbie madmartigano's Avatar
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    [SOLVED] how to start integrating e^(-x)*tan(e^(-x))dx

    I don't know how to begin with this problem:

    <br />
\int {{e^{ - x}}\tan {e^{ - x}}} dx<br />

    I've tried u substitution using {u = {e^{ - x}}}<br />
as well as integrations by parts using u = \tan ({e^{ - x}})<br />
and dv = {e^{ - x}}<br />
, but those only seem to complicate the problem.

    Any pointers?
    Last edited by madmartigano; March 7th 2009 at 12:41 AM.
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  2. #2
    Super Member
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    Quote Originally Posted by madmartigano View Post
    I've tried u substitution using {u = {e^{ - x}}}<br />
..., but those only seem to complicate the problem.
    I would beg to differ on that. Your substitution should make things clearer.

    u = e^{-x} \implies du = -e^{-x}~dx

    \int {{e^{ - x}}\tan {e^{ - x}}} dx = -\int \tan{u}~du = -\int \frac{\sin{u}}{\cos{u}}~du

    which should be straightforward.

    Alternatively, you can multiply by \frac{\sec{u}}{\sec{u}}.
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  3. #3
    MHF Contributor
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    Quote Originally Posted by madmartigano View Post
    I don't know how to begin with this problem:

    <br />
\int {{e^{ - x}}\tan {e^{ - x}}} dx<br />

    I've tried u substitution using {u = {e^{ - x}}}<br />
as well as integrations by parts using u = \tan ({e^{ - x}})<br />
and dv = {e^{ - x}}<br />
, but those only seem to complicate the problem.

    Any pointers?
    Hi

    The substitution u=e^{-x} works well
    du=-e^{-x}dx

    \int {{e^{ - x}}\tan {e^{ - x}}} dx = -\int \tan u\: du

    EDIT : too late !
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  4. #4
    Newbie madmartigano's Avatar
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    Heh... I miss the obvious too often. Thank you.
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