Question

`int(3x^2)/sqrt(9-16x^6)dx`
A. `(1)/(4)sin^(-1)((x^(3))/(3))+C`
B. `(1)/(4)sin^(-1)((4x^(3))/(3))+C`
C. `4sin^(-1)((x^(3))/(4))+C`
D. None of these

Answer 1

Correct Answer - B
Put `x^(3)=t and 3x^(2)dx=dt. `
`therefore I=int(dt)/(sqrt(9-16t^(2)))=(1)/(4)int (dt)/(sqrt((9)/(16)-t^(2)))=(1)/(4).int(dt)/(sqrt(((3)/(4))^(2)-t^(2)))`
`=(1)/(4)sin^(-1) ""(t)/(((3)/(4)))+C=(1)/(4)sin^(-1)((4x^(3))/(3))+C.`