मराठी

Evaluate the definite integral: ∫0π(sin2 x2-cos2 x2)dx

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प्रश्न

Evaluate the definite integral:

`int_0^pi (sin^2  x/2 - cos^2  x/2) dx`

बेरीज
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उत्तर

∴ `int_0^pi  (sin^2  x/2 - cos^2  x/2)  dx    ...(because cos^2 x/2 - sin^2 x = cos x)`

`= -int_0^pi  cos x  dx = - [sin x]_0^pi`

`= - (sin pi - sin 0)`

`= [0 - 0] = 0`

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पाठ 7: Integrals - Exercise 7.9 [पृष्ठ ३३८]

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एनसीईआरटी Mathematics Part 1 and 2 [English] Class 12
पाठ 7 Integrals
Exercise 7.9 | Q 18 | पृष्ठ ३३८

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