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Math Help - Is this done correctly?

  1. #1
    Junior Member
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    Is this done correctly?

    = \int [sec(\frac{x}{2}) + tan(\frac{3x+1}{5} )+ tan(7x)] dx

    = \int sec(\frac{x}{2}) + \int tan(\frac{3x+1}{5} ) + \int tan(7x) dx

    = 2ln|sec\frac{x}{2} + tan\frac{x}{2}|  + \int tan(\frac{3x+1}{5} )+ \int tan(7x) dx

    = 2ln|sec\frac{x}{2} + tan\frac{x}{2}|  + (\frac{5}{3})(-ln|cos \frac {3}{5} | )  + (\frac{1}{7})(-ln|cos 7|)

    = 2ln|sec\frac{x}{2} + tan\frac{x}{2}|   -\frac{5}{3}ln|cos \frac {3}{5} |   -\frac{1}{7}ln|cos 7|

    I don't have an answer to check but could someone give a thumbs up if they don't see anything wrong with it?
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  2. #2
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by circuscircus View Post
    = \int [sec(\frac{x}{2}) + tan(\frac{3x+1}{5} )+ tan(7x)] dx

    = \int sec(\frac{x}{2}) + \int tan(\frac{3x+1}{5} ) + \int tan(7x) dx

    = 2ln|sec\frac{x}{2} + tan\frac{x}{2}|  + \int tan(\frac{3x+1}{5} )+ \int tan(7x) dx

    = 2ln|sec\frac{x}{2} + tan\frac{x}{2}|  + (\frac{5}{3})(-ln|cos \frac {3}{5} | )  + (\frac{1}{7})(-ln|cos 7|)

    = 2ln|sec\frac{x}{2} + tan\frac{x}{2}|   -\frac{5}{3}ln|cos \frac {3}{5} |   -\frac{1}{7}ln|cos 7|

    I don't have an answer to check but could someone give a thumbs up if they don't see anything wrong with it?
    it should be 2 \ln \left| \sec \left( \frac {x}{2} \right) + \tan \left( \frac {x}{2} \right) \right| - \frac {5}{3} \ln \left| \cos \left( \frac {3x + 1}{5} \right) \right| - \frac {1}{7} \ln \left| \cos ( 7x ) \right| + C
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