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प्रश्न
Evaluate the definite integral:
`int_0^pi (sin^2 x/2 - cos^2 x/2) dx`
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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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