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Evaluate int(0)^(pi//2)lnsin2xdx...

Evaluate `int_(0)^(pi//2)lnsin2xdx`

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Let `I=int_(0)^(pi//2)lnsin2xdx` ltbr. `:'lnsin2((pi)/(2)-x)=ln sin(pi-2x)=lnsin2x`
`:.l=2int_(0)^(pi//4)lnsin2xdx` Put `2x=timpliesdx=(dt)/(2)`
`=int_(0)^(pi//2)lnsint dt`. Hence `l=int_(0)^(pi//2)ln sinxdx` (change of variables)…….`(i)` ltbr. `=int_90)^(pi//2)(lnsin((pi)/(2)-x)dx=int_(0)^(pi//2)ln cosx dx` .........`(ii)`
From `(i) + (ii)`, we have
`2I=int_(0)^(pi//2)ln(sinx*cosx)dx=int_(0)^(pi//2)lnsin2x-int_(0)^(pi//2)ln2dx`
`2I=l-(pi)/(2)` . So `I=-(pi)/(2)ln2`
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