Question #d453a

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Answer 1 · 2017-06-03T18:37:35

#int_0^(pi/2)ln(sin(x))dx= int_0^(pi/2)ln(cos(x))dx = int_0^(pi/2) ln(sin(2x))dx = I #

The last equality is more tricky you need to change variable with #x = 2u# and use parity at #pi/4#

#I = int_0^(pi/2)ln(sin(2x))dx#

#sin(2x) = 2cos(x)sin(x)#

#I = int_0^(pi/2)ln(2cos(x)sin(x))dx = int_0^(pi/2)ln(2)dx + int_0^(pi/2)ln(cos(x))dx + int_0^(pi/2)ln(sin(x))dx#

#I = int_0^(pi/2)ln(2)dx + I + I#

#-I = int_0^(pi/2)ln(2) dx#

#I = -pi/2ln(2)#