हिंदी

Evaluate the definite integral: ∫01dx1-x2

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प्रश्न

Evaluate the definite integral:

`int_0^1 dx/sqrt(1-x^2)`

योग
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उत्तर

`int_0^1  dx/sqrt(1 - x^2)`

`= [sin^-1 x]_0^1`

`= sin^-1 (1) - sin^-1 0`

= `pi/2 - 0`

`= pi/2`

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अध्याय 7: Integrals - Exercise 7.9 [पृष्ठ ३३८]

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एनसीईआरटी Mathematics Part 1 and 2 [English] Class 12
अध्याय 7 Integrals
Exercise 7.9 | Q 9 | पृष्ठ ३३८

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