The parametric form of a curve is x = t 3 t 2 - 1 ,   y = t t 2 - 1 , then ∫ d x x - 3 y =

The parametric form of a curve is x=t3t2-1, y=tt2-1, then ∫dxx-3y=
  1. 12logt2-1+C
  2. 2logtt2-1+C
  3. 14logtt2-3+C
  4. 52logt+1t2+C

Solution

Given x=t3t2-1, y=tt2-1

⇒x-3y=t3-3tt2-1

Also dx=t2-13t2-t32tt2-12dt=t4-3t2t2-12dt

Now, ∫dxx-3y=∫t4-3t2t2-12dtt3-3tt2-1=∫tt2-1dt

=12logt2-1+C

Asked in: AP EAMCET 2022 (04 Jul Shift 1)

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