`int_(0)^((pi)/(2))(sqrt(sinx))/(sqrt(sinx)+sqrt(cosx))dx`

Explanation

माना `I=int_(0)^(pi//2)(sqrt(sinx))/(sqrt(sinx)+sqrt(cosx))dx" …(1)"`
तब,`" "I=int_(0)^(pi//2)(sqrt(sin((pi)/(2)-x)))/(sqrt(sin((pi)/(2)-x))+sqrt(cos((pi)/(2)-x)))dx`
`" "[because int_(0)^(a)f(x)dx=int_(0)^(a)f(a-x)dx]`
`rArr" "I=int_(0)^(pi//2)(sqrt(cosx))/(sqrt(cosx)+sqrt(Sinx))dx" ...(2)"`
समीकरण (1 ) और (2 ) को जोड़ने पर ,
`2I=int_(0)^(pi//2)((sinx)+sqrt(cosx))/(sqrt(sinx)+sqrt(sinx))dx`
`=int_(0)^(pi//2)1dx=[x]_(0)^(pi//2)=(pi)/(2)`