Question

`int_0^1 (log x)/(sqrt(1-x^2))dx`

Answer 1

`x= sin theta,`when x=0; x=1,`theta= pi/2`
I=`int_(0)^(pi/2) (log(sin theta))/cos theta * cos theta d theta `
`= int_0^(pi/2) log (sin theta) d theta`
`I= int_0^(pi/2) log ( cos theta) d theta`
`2I= int_0^(pi/2) log (2 sin theta cos theta)/2 d theta`
`= int_0^(pi/2) [log(sin 2 theta) - log 2] d theta`
`= int_0^(pi/2) log(sin 2 theta) d theta - int_0^(pi/2) log (2) d theta`
`2 theta = y`
`2 dtheta = d y`
`= int_0^pi log(sin y )dy`
`= int_0^pi log(sin y) dy/2 - int_0^pi log(2) d theta`
`= 1/2 int_0^pi log ( sin y) dy - log (2) [ theta]`
`= 1/2 int_0^(pi/2) log(sin y ) dy + 1/2 int_(pi/2)^(pi) log(sin y ) dy - pi/2 log(2) `
`= 1/2 int_0^(pi/2) log(sin y ) dy + 1/2 int_0^(pi/2) log (cos z)dz - pi/2 log(2)`
`= 1/2 int_0^(pi/2) log(sin y ) dy + 1/2 int_0^(pi/2) log(sin z)dz - pi/2 log(2)`
`= 1/2 xx 2 int_0^(pi/2) log (sin y)dy - pi/2 log(2)`
`2I = 1 xx I - pi/2 log(2)`
`i = -pi/2 log(2)`
answer