How should I integrate 0 to pi/2 ln(sinx) dx

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$\int_0^{\frac{\pi}{2}} \ln{\sin{x}} dx$

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diviirockgod6

[minor flint]

Take the integral as I.
Do u = π/2 - x, then the only thing that changes is that the integral will be with cosine instead of sine. So, we get two representations:
I = ∫(ln(sin(x))dx, 0, π/2)
I = ∫(ln(cos(x))dx, 0, π/2)
So, we can add them:
2I = ∫(ln(sin(x)cos(x))dx, 0, π/2) = ∫((ln(sin(2x)) - ln(2))dx, 0, π/2) = ∫(ln(sin(2x))dx, 0, π/2) - (π/2)ln(2)
So:
I = (1/2)∫(ln(sin(2x))dx, 0, π/2) - (π/4)ln(2)
Next, do u = 2x, dx = (1/2)du.
I = (1/4)∫(ln(sin(u))du, 0, π) - (π/4)ln(2)
Next due to symmetry, we have ∫(ln(sin(u))du, 0, π/2) = ∫(ln(sin(u))du, π/2, π). So:
I = (1/2)∫(ln(sin(u))du, 0, π) - (π/4)ln(2)
And we recognize it as I again. So:
I = (1/2)I - (π/4)ln(2)
And now you can find I.

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-(π/2)ln(2)

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understood perfectly

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Yup.

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thanks

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Great! You're welcome 😁

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