If `int sqrt(1-x^2)/x^4dx=A(x).(sqrt(1-x^2))^m` where `A(x)` is a function of `x` then `(A(x))^m`= (A) `-1/(27x^9)` (B) `1/(27x)^9` (C) `1/(3x^9)` (D) `-1/(3x^9)`

Options

A. `(1)/(9x^(4))`
B. `(-1)/(3x^(3))`
C. `(-1)/(27x^(9))`
D. `(1)/(27x^(0))`

Answer

C

Explanation

....(i) We have, `int sqrt(1-x^(2))/(x^(4))dx = A(x)(sqrt(1-x^(2))^(m)+ C`
Let `I = int sqrt(1-x^(2))/(x^(4))dx = int sqrt(x^(2)((1)/(x^(2))-1))/x^(4)dx`
`= int (x sqrt((1)/(x^(2))-1))/(x^(4))dx = int (1)/(x^(3))sqrt((1)/(x^(2))-1)dx`
`=int(x sqrt((1)/(x^(2))-1))/(x^(4))dx=int(1)/(x^(3))sqrt((1)/(x^(2))-1)dx`
Put `(1)/(x^(2))-1 = t^(2) rArr (-2)/(x^(3))dx = 2t dt rArr (1)/(x^(3))dx = - t dt`
`therefore I = -intt^(2)dt = -(t^(3))/(3)+C`
`= -(1)/(3)((1-x^(2))/(x^(2)))^(3//2)+C [therefore t = ((1)/(x^(2))-1)^(1//2)]`
...(ii) `= -(1)/(3)(1)/(x^(3))(sqrt(1-x^(2)))^(3)+C`
On comparing Eqs. (i) and (ii), we get `A(x) = -(1)/(3x^(3)) and m = 3`
`therefore (A(x))^(m) = (A(x))^(3) = -(1)/(27x^(9))`