AP EAMCET20224 Jul 2022Morning ShiftMathematicsIndefinite IntegrationActual
The parametric form of a curve is x = t 3 t 2 - 1 , y = t t 2 - 1 , then ∫ d x x - 3 y =
Options
- A1 2 log t 2 - 1 + C
- B2 log t t 2 - 1 + C
- C1 4 log t t 2 - 3 + C
- D5 2 log t + 1 t 2 + C
Correct answer
A. 1 2 log t 2 - 1 + C
Step-by-step solution
Given x = t 3 t 2 - 1 ,   y = t t 2 - 1 ⇒ x - 3 y = t 3 - 3 t t 2 - 1 Also d x = t 2 - 1 3 t 2 - t 3 2 t t 2 - 1 2 d t = t 4 - 3 t 2 t 2 - 1 2 d t Now, ∫ d x x - 3 y = ∫ t 4 - 3 t 2 t 2 - 1 2 d t t 3 - 3 t t 2 - 1 = ∫ t t 2 - 1 d t = 1 2 log t 2 - 1 + C