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Math Help - U-substitution check

  1. #1
    Junior Member
    Joined
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    Exclamation U-substitution check

    Problem:
    \int\frac{sin(x)}{cos(x)}*dx from 0 to \pi/4

    Attempt:
    u = cos(x)
    du = -sin(x) dx
    dx = \frac{du}{-sin(x)}
    \int\frac{sin(x)}{-sin(x)*u}*du

    -\int\frac{du}{u}

    -\int\frac{du}{u} = -ln|u|

    -ln|cos(x)|

    -ln|cos(\pi/4)| + ln|cos(0)|


    My Answer:
    -ln(\frac{\sqrt{2}}{2})

    Correct Answer :
     <br />
\frac{ln|2|}{2}<br />

    Can someone tell me where I went wrong?
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  2. #2
    Super Member redsoxfan325's Avatar
    Joined
    Feb 2009
    From
    Swampscott, MA
    Posts
    943
    You didn't do anything wrong. They're the same answer.

    -\ln\left|\frac{\sqrt{2}}{2}\right| = -\ln\left|\frac{1}{\sqrt{2}}\right| = -\ln\left|2^{-1/2}\right| = -\frac{1}{2}\cdot-\ln|2| = \frac{\ln|2|}{2}
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  3. #3
    Member
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    Apr 2009
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    Edit: Might as well have read the whole post. -ln|\frac {\sqrt 2}{2}| = ln|\sqrt 2| = 1/2ln|2|

    \int \frac{sinx}{cosx}dx

    u = cosx
    du = -sinxdx

    \int \frac{du}{u} = ln|u|

    -\int \frac{du}{u} = -ln|cosx|
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