What is $int_(pi/2)^pi lnsinx$ ?

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Answer 1 · 2016-12-04T01:19:18

$int_(pi/2)^pi ln(sin x)dx = pi/2ln(1/2)$

Here you can see how to calculate that:

$int_0^(pi/2) ln(sin x)dx = pi/2ln(1/2)$

To calculate:

$int_(pi/2)^pi ln(sin x)dx$

just substitute:

$t=pi-x$

and you have:

$int_(pi/2)^pi ln(sin x)dx = int_(pi/2)^0 ln(sin (pi-t))(-dt)=int_0^(pi/2) ln(sint)dt$

because:

$sin(pi-t) = sin t$

and for the properties of the definite integral:

$int_a^b f(x)dx = -int_b^a f(x)dx$

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