`int_0^1sin^2 4x\ dx`

We recall that

`2\ sin^2theta=1-cos\ 2theta`

In this case, if `θ = 4x`,

`2 sin^2 4x` `=1-cos [2(4x)]` `=1-cos\ 8x`

So `sin^2 4x=(1-cos\ 8x)/2`

`int_0^1sin^2 4x dx =1/2int_0^1(1-cos\ 8x) dx`

`=1/2[x-(sin\ 8x)/8]_0^1`

`=1/2[(1-0.9894/8)-(0-0)]`

`=0.4382`

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