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#