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Math Help - Integral

  1. #1
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    Integral

    Show that \int_0^{\pi/2}\ln(\tan x)\,dx=0
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  2. #2
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by liyi View Post
    Show that \int_0^{\pi/2}\ln(\tan x)\,dx=0
    well, \int \ln (\tan x)~dx = \int \ln \left(  \frac {\sin x }{\cos x} \right)~dx = \int \ln (\sin x ) ~dx - \int \ln (\cos x)~dx

    by symmetry, the area "under" the curve \ln (\sin x) is the same as the area "under" the curve of \ln (\cos x) between the limits 0 and \frac {\pi}2. thus,

    \int_0^{\frac {\pi}2} \ln (\sin x)~dx = \int_0^{\frac {\pi}2} \ln (\cos x )~dx

    \Rightarrow \int_0^{\frac {\pi}2} \ln (\sin x)~dx - \int_0^{\frac {\pi}2} \ln (\cos x)~dx = 0 = \int_0^{\frac {\pi}2} \ln (\tan x )~dx
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  3. #3
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    Or we can say

    u=\int_0^{\pi /2} {\ln (\sin x)\,dx} .

    So u=\int_0^{\pi /2} {\ln (\cos x)\,dx} .

    (This last it's easy to prove with a simple substitution.)

    Finally

    \int_0^{\pi /2} {\ln (\sin x)\,dx} - \int_0^{\pi /2} {\ln (\cos x)\,dx} =\int_0^{\pi /2} {\ln (\tan x)\,dx} = 0\,\blacksquare
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  4. #4
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    A more interesting problem is so actually find,
    \int_0^{\pi/2} \ln (\sin x) dx.
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  5. #5
    Math Engineering Student
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    Quote Originally Posted by ThePerfectHacker View Post
    \int_0^{\pi/2} \ln (\sin x) dx.
    I agree. I like this integral

    Define \varphi = \int_0^{\pi /2} {\ln (\sin x)\,dx} .

    Set u=\frac\pi2-x,

    \varphi = \int_0^{\pi /2} {\ln (\sin x)\,dx} = \int_0^{\pi /2} {\ln (\cos x)\,dx} .

    Now adding this integrals,

    2\varphi = \int_0^{\pi /2} {\ln (\sin 2x)\,dx} - \frac{\pi }{2}\ln 2.

    For the last integral, make a substitution according to \alpha=2x,

    2\varphi = \frac{1}{2}\int_0^\pi {\ln (\sin \alpha )\,d\alpha } - \frac{\pi }{2}\ln 2.

    Since \int_0^\pi {\ln (\sin \alpha )\,d\alpha } = \int_0^{\pi /2} {\ln (\sin \alpha )\,d\alpha } + \int_{\pi /2}^\pi {\ln (\sin \alpha )\,d\alpha } ,

    let's make a final substitution defined by \beta=\pi-\alpha for the second integral, so

    \int_0^\pi {\ln (\sin \alpha )\,d\alpha } = 2\int_0^{\pi /2} {\ln (\sin \alpha )\,d\alpha } = 2\varphi .

    Therefore

    2\varphi = \varphi - \frac{\pi }{2}\ln 2\,\therefore \,\varphi = - \frac{\pi }{2}\ln 2\,\blacksquare
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