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Math Help - Need help evaluating this integral :S

  1. #1
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    Need help evaluating this integral :S



    i cant reealy hit the answer its too hard:
    this is what the back of the book says
    answer key:
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  2. #2
    Senior Member ecMathGeek's Avatar
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    Quote Originally Posted by ^_^Engineer_Adam^_^ View Post


    i cant reealy hit the answer its too hard:
    this is what the back of the book says
    answer key:
    You might be making this integration harder than it is.

    INT (2 - 3sin(2x))/cos(2x) dx
    2*INT sec(2x) dx - 3*INT tan(2x) dx

    We are now integrating sec(2x) and tan(2x) (I broke up the fraction and simplified the terms). For each, we will need a u substitution:
    Let u = 2x <--> du = 2 dx --> dx = 1/2 du
    2(1/2)*INT secu du - 3(1/2)*INT tanu du

    There are two basic formulas you can plug these integrations into. If you need them further explained, I can.
    INT secu du = ln|secu + tanu|
    INT tanu du = -ln|cosu|

    Therefore, we get:
    ln|sec(2x) + tan(2x)| + 3/2*ln|cos(2x)| + C

    As I'm sure you noticed, this does not match the answer in the back of the book, but with a little trig and log manipulation, it will.

    ln|sec(2x) + tan(2x)| = ln|1/cos(2x) + sin(2x)/cos(2x)| = ln|(1 + sin(2x))/cos(2x)| = ln|1 + sin(2x)| - ln|cos(2x)|

    Therefore, we get:
    ln|sec(2x) + tan(2x)| + 3/2*ln|cos(2x)| + C = ln|1 + sin(2x)| - ln|cos(2x)| + 3/2*ln|cos(2x)| + C = ln|1 + sin(2x)| + 1/2*ln|cos(2x)| + C
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