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

1.	`int dt/sqrt[3t-2t^2]`
Answer» `I = int dt/sqrt(3t-2t^2)` `3t-2t^2 = -(2t^2-2sqrt23/(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` We have, `sqrt2t-3/sqrt2 = x=>2t-3 = sqrt2x` `:. I = 1/sqrt2sin^-1((2(2t-3))/3)+c`

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