Integrate log(sinx) from 0 to pi /2?

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Answer 1 · 2017-09-26T08:37:54

$I=int_0^(pi/2)logsinxdx=-(pi/2)log2$

We use the property $int_0^af(x)dx=int_0^af(a-x)dx$

hence we can write $I=int_0^(pi/2)logsinxdx=int_0^(pi/2)logsin(pi/2-x)dx$

or $I=int_0^(pi/2)logsinxdx=int_0^(pi/2)logcosxdx$

or $2I=int_0^(pi/2)(logsinx+logcosx)dx=int_0^(pi/2)log(sinxcosx)dx$

= $int_0^(pi/2)log((sin2x)/2)dx=int_0^(pi/2)(logsin2x-log2)dx$

= $int_0^(pi/2)logsin2xdx-int_0^(pi/2)log2dx$

= $int_0^(pi/2)logsin2xdx-(pi/2)log2$ .............(A)

Let $I_1=int_0^(pi/2)logsin2xdx$ and $t=2x$ , then $I_1=1/2int_0^pilogsintdt$

and using the property $int_0^(2a)f(x)dx=2int_0^af(a-x)dx$ , if $f(2a-x)=f(x)$ - note that here $logsint=logsin(pi-t)$ and we get

$I_1=1/2int_0^pilogsintdt=int_0^(pi/2)logsintdt=I$

Hence (A) becomes $2I=I-(pi/2)log2$

or $I=-(pi/2)log2$