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D∫0π21-sin2x dx is equal to ______. - Mathematics

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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)`

MCQ
रिक्त स्थान भरें
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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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अध्याय 7: Integrals - Exercise [पृष्ठ १६९]

APPEARS IN

एनसीईआरटी एक्झांप्लर Mathematics [English] Class 12
अध्याय 7 Integrals
Exercise | Q 58 | पृष्ठ १६९

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