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Evaluate int0^(pi/2) sin^2x dx...

Evaluate `int_0^(pi/2) sin^2x dx`

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`int_0^(pi/2) sin^2x dx`
`=1/2int_0^(pi/2)2 sin^2x dx`
We know that
`cos2x=1−2sin^2x`
`=>2sin^2x=1-cos2x`
`=1/2int_0^(pi/2)(1-cos2x)dx`
`=1/2int_0^(pi/2)dx-1/2int_0^(pi/2)cos2x dx`
`=1/2[x]_0^(pi/2)-1/4 [-sin2x]_0^(pi/2)`
`=1/2[pi/2-0]+1/4[sin pi-sin 0]`
`=pi/4+0`
`=pi/4`
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