`int_(0)^(1)sin^(-1)((2x)/(1+x^(2)))dx`

Explanation

माना `I=int_(0)^(-1)sin^(-1)((2x)/(1+x^(2)))dx`
माना `tan theta=x" "rArr" "sec^(2) theta dx = dx`
`" "x=0" "rArr" "theta=0`
और `" "x=1" "rArr" "theta=(pi)/(4)`
`therefore" "I=int_(0)^(pi//4)sin^(-1)((2 tan theta)/(1+tan^(2)theta))sec^(2) theta d theta`
`=int_(0)^(pi//4)sin^(-1)(sin 2 theta)sec^(2) theta d theta`
`=2int_(0)^(pi//4)underset("I")(theta)underset(" II")(sec^(2))theta d theta`
`I=2[ theta int sec^(2) theta d theta - int{((d)/(d theta)theta)}int sec^(2) theta d theta]_(0)^(pi//4)`
`2[theta tan theta - int tan theta d theta ]_(0)^(pi//4)`
`=2[ theta tan theta+log |cos theta|]_(0)^(pi//4)`
`=2{((pi)/(4)tan.(pi)/(4)+log|cos.(pi)/(4)|)-(0+log|cos0|)}" "(because log 1=0)`
`=(pi)/(2)+2log(1)/(sqrt2)-0`
`=(pi)/(2)+2log2^(-1//2)` ltbr `=(pi)/(2)-2xx(1)/(2)log2" "(because log mn = n log m)`
`=(pi)/(2)-log 2`