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int dt/sqrt[3t-2t^2]...

`int dt/sqrt[3t-2t^2]`

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`I = int dt/sqrt(3t-2t^2)`
`3t-2t^2 = -(2t^2-2*sqrt2*3/(2sqrt2)t +(3/(2sqrt2))^2-9/8)`
`=-(sqrt2t-3/(2sqrt2))^2 + 9/8`
`=> I = int dt/(sqrt((3/(2sqrt2))^2-(sqrt2t-3/(2sqrt2))^2)`
Let `sqrt2t-3/(2sqrt2) = x=>sqrt2dt = dx`
`:. I = 1/sqrt2 int dx/(sqrt((3/(2sqrt2))^2-x^2`
we know, ` int 1/sqrt(a^2-x^2) = sin^-1(x/a)`
`:.I = 1/sqrt2sin^-1((2sqrt2x)/3)+c`
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