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[" 88."int(0)^( pi/2)log(e)(sin2x)dx=],[...

[" 88."int_(0)^( pi/2)log_(e)(sin2x)dx=],[" X) "1)pi log2],[" -12."f(pi log2quad 3)-(pi)/(2)log2quad " 4) "(pi)/(2)log2],[" 12."3(pi)/(2)log2quad " fourcr tum "],[[" () "2,-(pi)/(2)log2]]

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int_(0)^((pi)/(2))log(sin2x)dx

[int_(-pi/4)^( pi/4)log(cos x+sin x)dx=(1)/(2)],[[" 1) "pi log2," 2) "-pi log2," 3) "-(pi)/(4)log2," 4) "-(pi)/(2)log2]]

Show that int_(0)^(pi//2) log (sin x) dx = - pi/2 log 2

int_(0)^(pi)log(1+cosx)dx=-pi(log2)

int_(0)^(pi//2)(2logcosx-logsin2x)\ dx=-(pi)/(2)log2

int_(0)^(pi//2)log(tanx+cotx)dx=pi(log2)

int_(0)^((pi)/(2))(2log cos x-log sin2x)dx

Show that int_(0)^((pi)/(2))log(sin2x)dx=-(pi)/(2)(log2)

Show that int_(0)^((pi)/(2))logsinxdx=(pi)/(2)log((1)/(2))=(-pi)/(2)log2