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प्रश्न
`int_0^(pi/2) sqrt(1 - sin2x) "d"x` is equal to ______.
विकल्प
`2sqrt(2)`
`2(sqrt(2) + 1)`
2
`2(sqrt(2) - 1)`
उत्तर
`int_0^(pi/2) sqrt(1 - sin2x) "d"x` is equal to `2(sqrt(2) - 1)`.
Explanation:
Let 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)`.
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