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Thread: U-substitution check

  1. #1
    Junior Member
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    Exclamation U-substitution check

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

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

    $\displaystyle -\int\frac{du}{u}$

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

    $\displaystyle -ln|cos(x)|$

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


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

    Correct Answer :
    $\displaystyle
    \frac{ln|2|}{2}
    $

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

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

    $\displaystyle \int \frac{sinx}{cosx}dx$

    u = cosx
    du = -sinxdx

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

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