Results 1 to 5 of 5

Math Help - Int. prob

  1. #1
    Member javax's Avatar
    Joined
    Jan 2008
    From
    Milky Way
    Posts
    139

    Int. prob

    Hello, need some help on this integral!

    \int{\frac{1+sinx}{1+cosx}e^{x}dx}

    Thanks
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Senior Member
    Joined
    Feb 2008
    Posts
    410
    Quote Originally Posted by javax View Post
    Hello, need some help on this integral!

    \int{\frac{1+sinx}{1+cosx}e^{x}dx}

    Thanks
    Half-angle identities are key in many trig problems. Namely:

    \sin{x}=2\sin{\frac{x}{2}}\cos{\frac{x}{2}}

    \cos{x}=2\cos^2{\frac{x}{2}}-1

    Back to our integral:

    \int\frac{1+sinx}{1+cosx}e^{x}dx

    Substituting:

    \int\frac{1+2\sin{\frac{x}{2}}\cos{\frac{x}{2}}}{1  +2\cos^2{\frac{x}{2}}-1}e^{x}dx

    Simplifying:

    \int\frac{1+2\sin{\frac{x}{2}}\cos{\frac{x}{2}}}{2  \cos^2{\frac{x}{2}}}e^{x}dx

    \frac{1}{2}\int(\cos^{-2}{\frac{x}{2}})e^xdx+ \int(\tan{\frac{x}{2}})e^{x}dx

    \int(\tan{\frac{x}{2}})e^{x}dx+\int(\frac{1}{2}\se  c^{2}{\frac{x}{2}})e^xdx

    Let's part out the second integral:

    \int u\,dv=uv-\int v\,du

    Let u=e^x and dv=\frac{1}{2}\sec^2\frac{x}{2}dx

    Then:

    du=e^xdx

    v=\tan{\frac{x}{2}}

    Plugging all this in, we get:

    \int(\tan{\frac{x}{2}})e^{x}dx+\int(\frac{1}{2}\se  c^{2}{\frac{x}{2}})e^xdx=\int(\tan{\frac{x}{2}})e^  {x}dx+e^x\tan{\frac{x}{2}}-\int(\tan{\frac{x}{2}})e^xdx

    So:

    \int{\frac{1+sinx}{1+cosx}e^{x}dx}=e^x\tan{\frac{x  }{2}}
    Last edited by hatsoff; September 19th 2008 at 05:12 PM.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    o_O
    o_O is offline
    Primero Espada
    o_O's Avatar
    Joined
    Mar 2008
    From
    Canada
    Posts
    1,407
    = \int  \left(\frac{e^x \sin x}{1 + \cos x} + \frac{e^{x}}{1+ \cos x} \ \right)dx

    Three things we'll use:
    \frac{\sin x}{1+\cos x} = \tan \left(\frac{x}{2}\right)
    \frac{1}{1 + \cos x} = \frac{1}{2} \frac{2}{1 + \cos x} = \frac{1}{2} \sec^2 \left(\frac{x}{2}\right)

    And finally: \frac{d}{dx} \tan \left(\frac{x}{2}\right) = \frac{1}{2} \sec^2 \left(\frac{x}{2}\right)


    So, in an unconventional way:
    = \int \left( e^x\frac{\sin x}{1+ \cos x} +  \frac{1}{2} e^x \frac{2}{1 + \cos x} \right)\ dx

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

     = \int \left[ f'(x)g(x) + f(x)g'(x)\right]  dx = f(x)g(x) = e^x \tan \left(\frac{x}{2}\right)
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Member javax's Avatar
    Joined
    Jan 2008
    From
    Milky Way
    Posts
    139
    Indeed, It seems I cannot integrate by parts this one
    \frac{1}{2}\int(\cos^{-2}{\frac{x}{2}})e^xdx

    It's just getting longer...or I'm missing something!
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Member javax's Avatar
    Joined
    Jan 2008
    From
    Milky Way
    Posts
    139
    Quote Originally Posted by o_O View Post
    = \int \left(\frac{e^x \sin x}{1 + \cos x} + \frac{e^{x}}{1+ \cos x} \ \right)dx

    Three things we'll use:
    \frac{\sin x}{1+\cos x} = \tan \left(\frac{x}{2}\right)
    \frac{1}{1 + \cos x} = \frac{1}{2} \frac{2}{1 + \cos x} = \frac{1}{2} \sec^2 \left(\frac{x}{2}\right)

    And finally: \frac{d}{dx} \tan \left(\frac{x}{2}\right) = \frac{1}{2} \sec^2 \left(\frac{x}{2}\right)


    So, in an unconventional way:
    = \int \left( e^x\frac{\sin x}{1+ \cos x} + \frac{1}{2} e^x \frac{2}{1 + \cos x} \right)\ dx

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

     = \int \left[ f'(x)g(x) + f(x)g'(x)\right] dx = f(x)g(x) = e^x \tan \left(\frac{x}{2}\right)
    Cool, I remember this from the course now...! I have to float on my notebook a bit more

    Thanks
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Replies: 2
    Last Post: April 12th 2011, 10:13 AM
  2. AP prob
    Posted in the Pre-Calculus Forum
    Replies: 1
    Last Post: May 6th 2010, 09:53 PM
  3. Prob-Value
    Posted in the Advanced Statistics Forum
    Replies: 0
    Last Post: August 25th 2009, 04:49 AM
  4. Log Prob.
    Posted in the Algebra Forum
    Replies: 3
    Last Post: February 3rd 2009, 04:14 PM
  5. prob
    Posted in the Calculus Forum
    Replies: 1
    Last Post: December 2nd 2007, 09:00 AM

Search Tags


/mathhelpforum @mathhelpforum