Question

`int_0^(pi/2) sqrt(1 - sin2x)  "d"x` बराबर है

Options

• `2sqrt(2)`

• `2(sqrt(2) + 1)`

• 2

• `2(sqrt(2) - 1)`

Answer 1

सही उत्तर  `underline(2(sqrt(2) - 1))` है।

व्याख्या:

मान लीजिए I = `int_0^(pi/2) sqrt(1 - sin2x)  "d"x`

= `int_0^(pi/2) sqrt((sin^2x + cos^2x - 2 sinx cosx))  "d"x`

= `int_0^(pi/2) sqrt((sinx - cosx)^2)  "d"x`

= `int_0^(pi/2) +- (sinx - cosx)  "d"x`

= `int_0^(pi/4) - (sin x - cosx)  "d"x + int_(pi/4)^(pi/2) (sinx - cosx)  "dx`

= `int_0^(pi/4) (cosx - sinx)  "d"x + int_(pi/4)^(pi/2) (sinx - cosx)  "d"x`

= `[sinx + cosx]_0^(pi/4) + [- cosx - sinx]_(pi/4)^(pi/2)`

= `[(sin  pi/4 + cos  pi/4) - (sin0 - cos0)] - [(cos  pi/2 + sin  pi/2) - (cos  pi/4 + sin  pi/4)]`

= `[(1/sqrt(2) + 1/sqrt(2)) - (+ 1)] - [(0 + 1) - (1/sqrt(2) + 1/sqrt(2))]`

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

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

= `4/sqrt(2) - 2`

= `2sqrt(2) - 2`

= `2(sqrt(2) - 1)`.