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Evaluate the following: dttt∫dt3t-2t2 - Mathematics

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Question

Evaluate the following:

`int "dt"/sqrt(3"t" - 2"t"^2)`

Sum

Solution

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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Chapter 7: Integrals - Exercise [Page 164]

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NCERT Exemplar Mathematics [English] Class 12
Chapter 7 Integrals
Exercise | Q 15 | Page 164

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