Evaluate `int_0^(pi/2) logsinxdx`

Explanation

Let `I=int_(0)^(pi//2)log sin xdx`
Using the property,
`int_(0)^(2a)f(x)dx=int_(0)^(a)[f(x)=f(2a-x)]dx,` we get
`I=int_(0)^(pi//4)[logsin x+log sin ((pi)/(2)-x)]dx`
`=int_(0)^(pi//4)(log sin x+log cosx)dx`
`=int_(0)^(pi//4)log sin x cos x dx =int_(0)^(pi//4)log((2 sin x cosx)/(2))dx`
`=int_(0)^(pi//4)(log sin 2x-log2)dx=int_(0)^(pi//4)log sin 2x dx-int_(0)^(pi//4)log 2 dx`
`=I_(1)-I_(2)" ...(Say)" `
`I_(2)=int_(0)^(pi//4)log 2 dx=log 2 int_(0)^(pi//4)1dx`
`=log 2[x]_(0)^(pi//4)=(log2)[(pi)/(4)-0]=(pi)/(4)log2`
`I_(1)=int_(0)^(pi//4)log sin 2xdx`
Put `2x=t.` Then `dx=(dt)/(2)`
When `x=0, t=0`
When `x=pi//4, t=2((pi)/(4))=(pi)/(2)`
`therefore I_(1)=int_(0)^(pi//2)log sin txx(dt)/(2)`
`=(1)/(2)int_(0)^(pi//2)log sin x dx=(1)/(2)I" ..."[because int_(a)^(b)f(x)dx=int_(a)^(b)f(t)dt]`
`therefore I=(1)/(2)I-(pi)/(4)log2`
`therefore (1)/(2)I=-(pi)/(4)log2" "therefore I=-(pi)/(2)log 2=(pi)/(2)log((1)/(2)).`