Question

दर्शाइए कि  `int_0^(pi/2) (sin^2x)/(sinx + cosx) = 1/sqrt(2) log (sqrt(2) + 1)`

Answer 1

मान लीजिए I = `int_0^(pi/2) (sin^2x)/(sinx + cosx)  "d"x`

= `int_0^(pi/2) (sin^2(pi/2 - x))/(sin(pi/2 - x) + cos(pi/2 - x)) "d"x`  ....(P4 द्वारा)

⇒ I = `int_0^(pi/2) (cos^2x)/(sinx + cosx) "d"x`

अत:, हमें प्राप्त होता है: 2I = `1/sqrt(2)  int_0^(pi/2)  ("d"x)/(cos(x - pi/4))`

= `1/sqrt(2) int_0^(pi/2) sec(x - pi/2) "d"x`

= `1/sqrt(2) [log(sec(x - pi/4) + tan(x - pi/4))]_0^(pi/2)`

= `1/sqrt(2)[log(sec  pi/4 + tan  pi/4) - log sec(- pi/4) + tan(- pi/4)]`

= `1/sqrt(2) [log(sqrt(2) + 1) - log(sqrt(2) - 1)]`

= `1/sqrt(2) log|(sqrt(2) + 1)/(sqrt(2) - 1)|`

= `1/sqrt(2) log((sqrt(2) - 1)^2/1)`

= `2/sqrt(2) log(sqrt(2) + 1)`

अत:, I = `1/sqrt(2) log(sqrt(2) + 1)`.