Evaluate: `int_0^(1//2)1/((1+x^2)sqrt(1-x^2))`

Answer

N/a

Explanation

Let `" " I =int_(0)^(1//2) (dx)/((1+ x^(2))sqrt(1-x^(2)))`
Put `" " x = sin theta`
`implies " " dx = cos theta d theta `
As `x to 0` , then `theta to 0`
and `x to (1)/(2)` , then `theta to (pi)/(6)`
`therefore " " I = int_(0)^(pi//6) (cos theta)/((1 + sin^(2)theta) costheta) d theta = int_(0)^(pi//6) (1)/(1 +sin^(2) theta ) d theta `
= `int_(0)^(pi//6) (1)/(cos^(2) theta (sec^(2)theta + tan^(2)theta)) d theta `
`= int_(0)^(pi//6) (sec^(2) theta)/(sec^(2) theta + tan^(2) theta)d theta`
`int_(0)^(pi//6)(sec^(2) theta)/(1 + tan^(2) theta + tan^(2) theta) d theta `
`int_(0)^(pi//2)(sec^(2)theta)/(1 + 2 tan^(2) theta ) d theta`
Again , put `tan theta = t `
`implies " " sec^(2) theta d theta = dt `
As `theta" to "0` then `t to 0`
and `theta to (pi)/(6)` , then `t to (1)/(sqrt3)`
`I = int_(0)^(1//sqrt3) (dt)/(1+ 2t^(2)) = (1)/(2)int_(0)^(1//sqrt3) (dt)/(((1)/(sqrt2))^(2) + t^(2))`
`= (1)/(2) * (1)/(1//sqrt2) [tan^(-1)"(t)/((1)/(sqrt2))]_(0)^(1//sqrt3) = (1)/(sqrt2) [tan^(-1) (sqrt2 t )]_(0)^(1//sqrt3)`
`=(1)/(sqrt2)[tan^(-1)sqrt((2)/(3)) - 0] = (1)/(sqrt2) tan^(-1) (sqrt((2)/(3)))`