If `y=int1/(1+x^2)^(3/2)dx` and `y=0` when `x=0` , then value of y when `x=1` is

Options

A. `(1)/(sqrt(2))`
B. `sqrt(2)`
C. `2sqrt(2)`
D. none of these

Answer

A

Explanation

`"Let " x=tan theta. " Then " dx=sec^(2)thetad theta. " Now,"`
` y=int(dx)/((1+x^(2))^((3)/(2)))=int (sec^(2)theta)/((1+tan^(2)theta)^((3)/(2)))d theta`
`=int(sec^(2)theta)/((sec^(2)theta)^((3)/(2)))d theta`
`=int(sec^(2)theta)/(sec^(3)theta)d theta=int(d theta)/(sec theta)=int cos theta d theta`
`"Hence, " y=sin theta+c=(x)/(sqrt(1+x^(2)))+c " " (1)`
`" " [:' tan theta=x=(x)/(1),sin theta=(x)/(sqrt(1^(2)+x^(2)))] `
`"Given when " x=0, y=0 " from equation(1),"`
`0=0+c " or " c=0 `
`"Thus, from equation(1), " y=(x)/(sqrt(1+x^(2))). `
`"Hence, when " x=1,y=(1)/(sqrt(2)).`