`int_(0)^((pi)/(2))(sinx)/(1+cos^(2)x)dx`

Explanation

माना `" "I=int_(0)^(pi//2)(sinx)/(1+cos^(2)x)dx`
माना `" "cos x=t`
`rArr -sin x dx=dt rArr sin x dx=- dt`
`x =0" "rArr" "t=cos 0=1`
और `" "x=(pi)/(2)" "rArr" "t=cos.(pi)/(2)=0`
`therefore" "I=int_(1)^(0)(1)/(1+t^(2))(-dt)=-int_(1)^(0)(1)/(1+t^(2))dt`
`=-[tan^(-1)t]_(1)^(0)`
`=-[tan^(-1)(0)-tan^(-1)(1)]`
`=-(0-(pi)/(4))=(pi)/(4)`