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    Anyone have any ideas on this one? (attached)

    Thanks,

    Vazzyb
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    I = int(0, pi/2) ln(sin(x)) dx

    Let u=pi/2-x, then du=-dx. So:
    I = int(pi/2, 0) ln(sin(pi/2 - u)) -du
    = - int(pi/2, 0) ln(cos(u)) du
    = int(0, pi/2) ln(cos(u)) du

    This gives us the first equality. Now ln(sin(2u)) = ln(2sin(u)cos(u)) = ln(2) + ln(sin(u)) + ln(cos(u)). So:
    I = int(0, pi/2) ln(sin(2u)) - ln(2) - ln(sin(u)) du
    = int(0, pi/2) ln(sin(2u)) du - int(0, pi/2) ln(2) du - int(0, pi/2) ln(sin(u)) du
    = int(0, pi/2) ln(sin(2u)) du - pi/2 ln(2) - I
    => 2I = int(0, pi/2) ln(sin(2u)) du - pi/2 ln(2)
    => I = (1/2) int(0, pi/2) ln(sin(2u)) du - pi/4 ln(2)

    And that gives us the second equality. Now let's move on to the second part.

    J = int(0, pi/2) ln(sin(2x)) dx

    Let v=2x, then dv=2dx. So:
    J = (1/2) int(0, pi) ln(sin(v)) dv

    This gives us the first equality. Let's go a step further:
    J = (1/2) int(0, pi) ln(sin(v)) dv
    = (1/2) int(0, pi/2) ln(sin(v)) dv + (1/2) int(pi/2, pi) ln(sin(v)) dv

    Let v=pi-u, then dv=-du. So:
    int(pi/2, pi) ln(sin(v)) dv = int(pi/2, 0) ln(sin(pi-u)) -du
    = - int(pi/2, 0) ln(sin(u)) du
    = int(0, pi/2) ln(sin(u)) du

    So let's sub this back into J:
    J = (1/2) int(0, pi/2) ln(sin(v)) dv + (1/2) int(pi/2, pi) ln(sin(v)) dv
    = (1/2) int(0, pi/2) ln(sin(v)) dv + (1/2) int(0, pi/2) ln(sin(u)) du
    = int(0, pi/2) ln(sin(x)) dx
    • Thread Starter
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    Thank you so much for that. Very helpful. Will rep tommorow.
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    No problem. Out of curiosity, which level of STEP is this?
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    Step 2.
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    Can I ask what year it was?
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    1991 Step 2.
 
 
 
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Updated: April 9, 2006
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