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प्रश्न
Evaluate the following:
`int "dt"/sqrt(3"t" - 2"t"^2)`
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उत्तर
Let I = `int "dt"/sqrt(3"t" - 2"t"^2)`
= `int "dt"/sqrt(-2("t"^2 - 3/2 "t"))`
= `1/sqrt(2) int "dt"/sqrt(-("t"^2 - 3/2 "t" + 9/16 - 9/16))` ....[Making perfect square]
= `1/sqrt(2) int "dt"/sqrt(-[("t" - 3/4)^2 - 9/16])`
= `1/sqrt(2) int "dt"/sqrt(9/16 - ("t" - 3/4)^2)`
= `1/sqrt(2) int "dt"/sqrt((3/4)^2 - ("t" - 3/4)^2)`
= `1/sqrt(2) * sin^-1 ("t" - 3/4)/(3/4) + "C"`
= `1/sqrt(2) sin^-1 (4"t" - 3)/3 + "C"`
Hence, I = `1/sqrt(2) sin^-1 ((4"t" - 3)/3) + "C"`
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