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प्रश्न
Evaluate: `int_0^pi cos^2 x "d"x`
उत्तर
`int_0^pi cos^2 x "d"x = int_0^pi((1 + cos 2x)/2) "d"x`
= `1/2[int_0^pi "d"x + int_0^pi cos 2x "d"x]`
= `1/2[[x]_0^pi + [(sin 2x)/2]_0^pi]`
= `1/2[(pi - 0) + 1/2(sin 2pi - sin 0)]`
= `1/2[pi + 1/2(0 - 0)]`
= `pi/2`
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